Antidifferentiation Review Integration (antidifferentiation) is the reverse process of differentiation. If F " ( x ) = f ( x ) F"(x) = f(x) F " ( x ) = f ( x )
∫ f ( x ) d x = F ( x ) + C \int f(x)\,dx = F(x) + C ∫ f ( x ) d x = F ( x ) + C Where C C C
Basic Integration Rules Rule Formula Power rule ∫ x n d x = x n + 1 n + 1 + C ( n ≠ − 1 ) \displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) ∫ x n d x = n + 1 x n + 1 + C ( n = − 1 ) Reciprocal $\displaystyle\int \frac{1}{x},dx = \ln|x| Constant multiple ∫ k f ( x ) d x = k ∫ f ( x ) d x \displaystyle\int kf(x)\,dx = k\int f(x)\,dx ∫ k f ( x ) d x = k ∫ f ( x ) d x Sum/difference ∫ [ f ( x ) ± g ( x ) ] d x = ∫ f ( x ) d x ± ∫ g ( x ) d x \displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx ∫ [ f ( x ) ± g ( x )] d x = ∫ f ( x ) d x ± ∫ g ( x ) d x
Standard Integrals f ( x ) f(x) f ( x ) ∫ f ( x ) d x \int f(x)\,dx ∫ f ( x ) d x e x e^x e x e x + C e^x + C e x + C a x a^x a x a x ln a + C \dfrac{a^x}{\ln a} + C ln a a x + C sin x \sin x sin x − cos x + C -\cos x + C − cos x + C cos x \cos x cos x sin x + C \sin x + C sin x + C sec 2 x \sec^2 x sec 2 x tan x + C \tan x + C tan x + C csc 2 x \csc^2 x csc 2 x − cot x + C -\cot x + C − cot x + C sec x tan x \sec x \tan x sec x tan x sec x + C \sec x + C sec x + C 1 1 − x 2 \dfrac{1}{\sqrt{1-x^2}} 1 − x 2 1 arcsin x + C \arcsin x + C arcsin x + C 1 1 + x 2 \dfrac{1}{1+x^2} 1 + x 2 1 arctan x + C \arctan x + C arctan x + C
The Fundamental Theorem of Calculus First Fundamental Theorem If f f f [ a , b ] [a, b] [ a , b ] F ′ ( x ) = f ( x ) F'(x) = f(x) F ′ ( x ) = f ( x )
∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) Second Fundamental Theorem If f f f a a a
\frac{d}`\{dx}`\left[\int_a^x f(t)\,dt\right] = f(x) Definite Integrals Properties For continuous functions f f f g g g [ a , b ] [a, b] [ a , b ]
∫ a b k f ( x ) d x = k ∫ a b f ( x ) d x \displaystyle\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx ∫ a b k f ( x ) d x = k ∫ a b f ( x ) d x ∫ a b [ f ( x ) ± g ( x ) ] d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x \displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx ∫ a b [ f ( x ) ± g ( x )] d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x ∫ a a f ( x ) d x = 0 \displaystyle\int_a^a f(x)\,dx = 0 ∫ a a f ( x ) d x = 0 ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x \displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x c c c Even and Odd Functions If f f f f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x )
∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x If f f f f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x )
∫ − a a f ( x ) d x = 0 \int_{-a}^{a} f(x)\,dx = 0 ∫ − a a f ( x ) d x = 0 Example
Evaluate ∫ − 2 2 x 3 cos ( x 2 ) d x \displaystyle\int_{-2}^{2} x^3\cos(x^2)\,dx ∫ − 2 2 x 3 cos ( x 2 ) d x
Since x 3 cos ( x 2 ) x^3\cos(x^2) x 3 cos ( x 2 ) × \times × = = =
∫ − 2 2 x 3 cos ( x 2 ) d x = 0 \int_{-2}^{2} x^3\cos(x^2)\,dx = 0 ∫ − 2 2 x 3 cos ( x 2 ) d x = 0 Integration by Substitution Method When the integrand contains a function and its derivative (or a composite function), substitute u = g ( x ) u = g(x) u = g ( x ) g ( x ) g(x) g ( x )
∫ f ( g ( x ) ) g ′ ( x ) d x = ∫ f ( u ) d u \int f(g(x))g'(x)\,dx = \int f(u)\,du ∫ f ( g ( x )) g ′ ( x ) d x = ∫ f ( u ) d u Steps Choose u = g ( x ) u = g(x) u = g ( x ) Compute d u d x = g ′ ( x ) \dfrac{du}{dx} = g'(x) d x d u = g ′ ( x ) d u = g ′ ( x ) d x du = g'(x)\,dx d u = g ′ ( x ) d x Rewrite the integral in terms of u u u Integrate with respect to u u u Substitute back u = g ( x ) u = g(x) u = g ( x ) Example
Evaluate ∫ 2 x x 2 + 1 d x \displaystyle\int 2x\sqrt{x^2+1}\,dx ∫ 2 x x 2 + 1 d x
Let u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 d u d x = 2 x \dfrac{du}{dx} = 2x d x d u = 2 x d u = 2 x d x du = 2x\,dx d u = 2 x d x
∫ u d u = ∫ u 1 / 2 d u = 2 3 u 3 / 2 + C = 2 3 ( x 2 + 1 ) 3 / 2 + C \int \sqrt{u}\,du = \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^2+1)^{3/2} + C ∫ u d u = ∫ u 1/2 d u = 3 2 u 3/2 + C = 3 2 ( x 2 + 1 ) 3/2 + C Example
Evaluate ∫ 0 1 x e x 2 d x \displaystyle\int_0^1 xe^{x^2}\,dx ∫ 0 1 x e x 2 d x
Let u = x 2 u = x^2 u = x 2 d u = 2 x d x du = 2x\,dx d u = 2 x d x
When x = 0 x = 0 x = 0 u = 0 u = 0 u = 0 x = 1 x = 1 x = 1 u = 1 u = 1 u = 1
∫ 0 1 x e x 2 d x = 1 2 ∫ 0 1 e u d u = 1 2 [ e u ] 0 1 = 1 2 ( e − 1 ) \int_0^1 xe^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}\big[e^u\big]_0^1 = \frac{1}{2}(e - 1) ∫ 0 1 x e x 2 d x = 2 1 ∫ 0 1 e u d u = 2 1 [ e u ] 0 1 = 2 1 ( e − 1 ) Trigonometric Substitutions Certain integrals can be simplified using trigonometric substitutions:
Expression Substitution Identity a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x = a sin θ x = a\sin\theta x = a sin θ 1 − sin 2 θ = cos 2 θ 1 - \sin^2\theta = \cos^2\theta 1 − sin 2 θ = cos 2 θ a 2 + x 2 \sqrt{a^2 + x^2} a 2 + x 2 x = a tan θ x = a\tan\theta x = a tan θ 1 + tan 2 θ = sec 2 θ 1 + \tan^2\theta = \sec^2\theta 1 + tan 2 θ = sec 2 θ x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2 x = a sec θ x = a\sec\theta x = a sec θ sec 2 θ − 1 = tan 2 θ \sec^2\theta - 1 = \tan^2\theta sec 2 θ − 1 = tan 2 θ
Example
Evaluate ∫ 1 4 − x 2 d x \displaystyle\int \frac{1}{\sqrt{4-x^2}}\,dx ∫ 4 − x 2 1 d x
This is a standard integral: arcsin ( x 2 ) + C \arcsin\!\left(\dfrac{x}{2}\right) + C arcsin ( 2 x ) + C
Alternatively, let x = 2 sin θ x = 2\sin\theta x = 2 sin θ d x = 2 cos θ d θ dx = 2\cos\theta\,d\theta d x = 2 cos θ d θ
∫ 2 cos θ 2 cos θ d θ = ∫ 1 d θ = θ + C = arcsin ( x 2 ) + C \int \frac{2\cos\theta}{2\cos\theta}\,d\theta = \int 1\,d\theta = \theta + C = \arcsin\!\left(\frac{x}{2}\right) + C ∫ 2 cos θ 2 cos θ d θ = ∫ 1 d θ = θ + C = arcsin ( 2 x ) + C Integration by Parts For the product of two functions:
∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u Choosing u u u d v dv d v Use the LIATE rule (prioritise choosing u u u
L ogarithmic functionsI nverse trig functionsA lgebraic functions (polynomials)T rigonometric functionsE xponential functionsExample
Evaluate ∫ x e x d x \displaystyle\int x e^x\,dx ∫ x e x d x
Let u = x u = x u = x d v = e x d x dv = e^x\,dx d v = e x d x
Then d u = d x du = dx d u = d x v = e x v = e^x v = e x
∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C \int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x - 1) + C ∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C Example
Evaluate ∫ x sin x d x \displaystyle\int x\sin x\,dx ∫ x sin x d x
Let u = x u = x u = x d v = sin x d x dv = \sin x\,dx d v = sin x d x
Then d u = d x du = dx d u = d x v = − cos x v = -\cos x v = − cos x
∫ x sin x d x = − x cos x − ∫ ( − cos x ) d x = − x cos x + sin x + C \int x\sin x\,dx = -x\cos x - \int (-\cos x)\,dx = -x\cos x + \sin x + C ∫ x sin x d x = − x cos x − ∫ ( − cos x ) d x = − x cos x + sin x + C Repeated Integration by Parts For integrals like ∫ x 2 e x d x \displaystyle\int x^2 e^x\,dx ∫ x 2 e x d x ∫ x 2 sin x d x \displaystyle\int x^2 \sin x\,dx ∫ x 2 sin x d x
Example
Evaluate ∫ x 2 e x d x \displaystyle\int x^2 e^x\,dx ∫ x 2 e x d x
First application: u = x^2$$dv = e^x\,dx :
∫ x 2 e x d x = x 2 e x − ∫ 2 x e x d x \int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx ∫ x 2 e x d x = x 2 e x − ∫ 2 x e x d x Second application on ∫ 2 x e x d x \int 2x e^x\,dx ∫ 2 x e x d x u = 2x$$dv = e^x\,dx :
∫ 2 x e x d x = 2 x e x − ∫ 2 e x d x = 2 x e x − 2 e x \int 2x e^x\,dx = 2xe^x - \int 2e^x\,dx = 2xe^x - 2e^x ∫ 2 x e x d x = 2 x e x − ∫ 2 e x d x = 2 x e x − 2 e x Combining:
∫ x 2 e x d x = x 2 e x − 2 x e x + 2 e x + C = e x ( x 2 − 2 x + 2 ) + C \int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C ∫ x 2 e x d x = x 2 e x − 2 x e x + 2 e x + C = e x ( x 2 − 2 x + 2 ) + C Cyclic Integration by Parts Integrals like ∫ e x sin x d x \displaystyle\int e^x \sin x\,dx ∫ e x sin x d x
Example
Evaluate ∫ e x sin x d x \displaystyle\int e^x \sin x\,dx ∫ e x sin x d x
Let u = e^x$$dv = \sin x\,dx . Then du = e^x\,dx$$v = -\cos x .
I = − e x cos x + ∫ e x cos x d x I = -e^x\cos x + \int e^x\cos x\,dx I = − e x cos x + ∫ e x cos x d x Now let u = e^x$$dv = \cos x\,dx . Then du = e^x\,dx$$v = \sin x .
∫ e x cos x d x = e x sin x − ∫ e x sin x d x = e x sin x − I \int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I ∫ e x cos x d x = e x sin x − ∫ e x sin x d x = e x sin x − I Substituting back:
I = − e x cos x + e x sin x − I I = -e^x\cos x + e^x\sin x - I I = − e x cos x + e x sin x − I 2 I = e x ( sin x − cos x ) 2I = e^x(\sin x - \cos x) 2 I = e x ( sin x − cos x ) I = e x ( sin x − cos x ) 2 + C I = \frac{e^x(\sin x - \cos x)}{2} + C I = 2 e x ( sin x − cos x ) + C Partial Fractions When the integrand is a rational function with a denominator that can be factorised, decompose into Partial fractions.
Types of Decomposition Denominator Partial Fractions ( a x + b ) ( c x + d ) (ax+b)(cx+d) ( a x + b ) ( c x + d ) A a x + b + B c x + d \dfrac{A}{ax+b} + \dfrac{B}{cx+d} a x + b A + c x + d B ( a x + b ) 2 (ax+b)^2 ( a x + b ) 2 A a x + b + B ( a x + b ) 2 \dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2} a x + b A + ( a x + b ) 2 B ( a x + b ) ( x 2 + c ) (ax+b)(x^2+c) ( a x + b ) ( x 2 + c ) A a x + b + B x + C x 2 + c \dfrac{A}{ax+b} + \dfrac{Bx+C}{x^2+c} a x + b A + x 2 + c B x + C
Example
Evaluate ∫ 5 x + 1 ( x + 1 ) ( x − 2 ) d x \displaystyle\int \frac{5x+1}{(x+1)(x-2)}\,dx ∫ ( x + 1 ) ( x − 2 ) 5 x + 1 d x
5 x + 1 ( x + 1 ) ( x − 2 ) = A x + 1 + B x − 2 \frac{5x+1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2} ( x + 1 ) ( x − 2 ) 5 x + 1 = x + 1 A + x − 2 B 5 x + 1 = A ( x − 2 ) + B ( x + 1 ) 5x + 1 = A(x-2) + B(x+1) 5 x + 1 = A ( x − 2 ) + B ( x + 1 ) x = − 1 x = -1 x = − 1 − 5 + 1 = A ( − 3 ) ⟹ A = 4 3 -5 + 1 = A(-3) \implies A = \dfrac{4}{3} − 5 + 1 = A ( − 3 ) ⟹ A = 3 4
x = 2 x = 2 x = 2 10 + 1 = B ( 3 ) ⟹ B = 11 3 10 + 1 = B(3) \implies B = \dfrac{11}{3} 10 + 1 = B ( 3 ) ⟹ B = 3 11
∫ 5 x + 1 ( x + 1 ) ( x − 2 ) d x = 4 3 ln ∥ x + 1 ∥ + 11 3 ln ∥ x − 2 ∥ + C \int \frac{5x+1}{(x+1)(x-2)}\,dx = \frac{4}{3}\ln\|x+1\| + \frac{11}{3}\ln\|x-2\| + C ∫ ( x + 1 ) ( x − 2 ) 5 x + 1 d x = 3 4 ln ∥ x + 1∥ + 3 11 ln ∥ x − 2∥ + C Area Under Curves Area Between a Curve and the x x x The signed area between y = f ( x ) y = f(x) y = f ( x ) x x x x = a x = a x = a x = b x = b x = b
A r e a = ∫ a b f ( x ) d x \mathrm{Area} = \int_a^b f(x)\,dx Area = ∫ a b f ( x ) d x If the curve crosses the x x x
Area Between Two Curves The area between y = f ( x ) y = f(x) y = f ( x ) y = g ( x ) y = g(x) y = g ( x ) x = a x = a x = a x = b x = b x = b f ( x ) ≥ g ( x ) f(x) \ge g(x) f ( x ) ≥ g ( x )
A r e a = ∫ a b [ f ( x ) − g ( x ) ] d x \mathrm{Area} = \int_a^b [f(x) - g(x)]\,dx Area = ∫ a b [ f ( x ) − g ( x )] d x Example
Find the area enclosed by y = x 2 y = x^2 y = x 2 y = 2 x y = 2x y = 2 x
Intersection points: x 2 = 2 x ⟹ x ( x − 2 ) = 0 ⟹ x = 0 x^2 = 2x \implies x(x-2) = 0 \implies x = 0 x 2 = 2 x ⟹ x ( x − 2 ) = 0 ⟹ x = 0 x = 2 x = 2 x = 2
For 0 ≤ x ≤ 2 0 \le x \le 2 0 ≤ x ≤ 2 2 x ≥ x 2 2x \ge x^2 2 x ≥ x 2
A r e a = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − x 3 3 ] 0 2 = 4 − 8 3 = 4 3 \mathrm{Area} = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} Area = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − 3 x 3 ] 0 2 = 4 − 3 8 = 3 4 Example
Find the area enclosed by y = x 2 − 4 y = x^2 - 4 y = x 2 − 4 x x x
Zeros: x 2 − 4 = 0 ⟹ x = ± 2 x^2 - 4 = 0 \implies x = \pm 2 x 2 − 4 = 0 ⟹ x = ± 2
A r e a = ∫ − 2 2 ∣ x 2 − 4 ∣ d x = 2 ∫ 0 2 ( 4 − x 2 ) d x = 2 [ 4 x − x 3 3 ] 0 2 = 2 ( 8 − 8 3 ) = 32 3 \mathrm{Area} = \int_{-2}^{2} |x^2 - 4|\,dx = 2\int_0^2 (4 - x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3} Area = ∫ − 2 2 ∣ x 2 − 4∣ d x = 2 ∫ 0 2 ( 4 − x 2 ) d x = 2 [ 4 x − 3 x 3 ] 0 2 = 2 ( 8 − 3 8 ) = 3 32 Volume of Revolution Rotation about the x x x The volume generated by rotating y = f ( x ) y = f(x) y = f ( x ) x = a x = a x = a x = b x = b x = b x x x
V = π ∫ a b [ f ( x ) ] 2 d x V = \pi \int_a^b [f(x)]^2\,dx V = π ∫ a b [ f ( x ) ] 2 d x Rotation about the y y y The volume generated by rotating x = g ( y ) x = g(y) x = g ( y ) y = c y = c y = c y = d y = d y = d y y y
V = π ∫ c d [ g ( y ) ] 2 d y V = \pi \int_c^d [g(y)]^2\,dy V = π ∫ c d [ g ( y ) ] 2 d y Alternatively, using the shell method about the y y y
V = 2 π ∫ a b x ∣ f ( x ) ∣ d x V = 2\pi \int_a^b x |f(x)|\,dx V = 2 π ∫ a b x ∣ f ( x ) ∣ d x Example
Find the volume generated by rotating the region bounded by y = \sqrt{x}$$x = 4 And the x x x x x x
V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ x 2 2 ] 0 4 = 8 π V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi V = π ∫ 0 4 ( x ) 2 d x = π ∫ 0 4 x d x = π [ 2 x 2 ] 0 4 = 8 π Example
Find the volume generated by rotating the region bounded by y = x^2$$y = 0 And x = 1 x = 1 x = 1 y y y
Using the disc method (in terms of y y y
x = y x = \sqrt{y} x = y y = 0 y = 0 y = 0 y = 1 y = 1 y = 1
V = π ∫ 0 1 ( y ) 2 d y = π ∫ 0 1 y d y = π [ y 2 2 ] 0 1 = π 2 V = \pi\int_0^1 (\sqrt{y})^2\,dy = \pi\int_0^1 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^1 = \frac{\pi}{2} V = π ∫ 0 1 ( y ) 2 d y = π ∫ 0 1 y d y = π [ 2 y 2 ] 0 1 = 2 π Volume between two curves When rotating the region between y = f ( x ) y = f(x) y = f ( x ) y = g ( x ) y = g(x) y = g ( x ) x x x
V = π ∫ a b ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x V = \pi\int_a^b \left([f(x)]^2 - [g(x)]^2\right)\,dx V = π ∫ a b ( [ f ( x ) ] 2 − [ g ( x ) ] 2 ) d x Example
Find the volume generated by rotating the region between y = x y = x y = x y = x 2 y = x^2 y = x 2 x x x
Intersection: x = x 2 ⟹ x = 0 x = x^2 \implies x = 0 x = x 2 ⟹ x = 0 x = 1 x = 1 x = 1
V = π ∫ 0 1 ( x 2 − x 4 ) d x = π [ x 3 3 − x 5 5 ] 0 1 = π ( 1 3 − 1 5 ) = 2 π 15 V = \pi\int_0^1 \left(x^2 - x^4\right)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15} V = π ∫ 0 1 ( x 2 − x 4 ) d x = π [ 3 x 3 − 5 x 5 ] 0 1 = π ( 3 1 − 5 1 ) = 15 2 π Kinematics Applications Displacement, Velocity, and Acceleration Velocity is the derivative of displacement: v = d s d t v = \dfrac{ds}{dt} v = d t d s Acceleration is the derivative of velocity: a = d v d t = d 2 s d t 2 a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2} a = d t d v = d t 2 d 2 s Displacement from velocity: s = ∫ v d t s = \displaystyle\int v\,dt s = ∫ v d t Velocity from acceleration: v = ∫ a d t v = \displaystyle\int a\,dt v = ∫ a d t Total Distance vs Displacement D i s p l a c e m e n t = ∫ t 1 t 2 v ( t ) d t \mathrm{Displacement} = \int_{t_1}^{t_2} v(t)\,dt Displacement = ∫ t 1 t 2 v ( t ) d t T o t a l d i s t a n c e = ∫ t 1 t 2 ∣ v ( t ) ∣ d t \mathrm{Total distance} = \int_{t_1}^{t_2} |v(t)|\,dt Totaldistance = ∫ t 1 t 2 ∣ v ( t ) ∣ d t Example
A particle moves with velocity v ( t ) = t 2 − 4 t + 3 m / s v(t) = t^2 - 4t + 3\mathrm{ m/s} v ( t ) = t 2 − 4 t + 3 m/s 0 ≤ t ≤ 5 0 \le t \le 5 0 ≤ t ≤ 5
(a) Find the displacement.
S = ∫ 0 5 ( t 2 − 4 t + 3 ) d t = [ t 3 3 − 2 t 2 + 3 t ] 0 5 = 125 3 − 50 + 15 = 40 3 m S = \int_0^5 (t^2 - 4t + 3)\,dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^5 = \frac{125}{3} - 50 + 15 = \frac{40}{3}\mathrm{ m} S = ∫ 0 5 ( t 2 − 4 t + 3 ) d t = [ 3 t 3 − 2 t 2 + 3 t ] 0 5 = 3 125 − 50 + 15 = 3 40 m (b) Find the total distance travelled.
Find when v = 0 v = 0 v = 0 t 2 − 4 t + 3 = 0 ⟹ ( t − 1 ) ( t − 3 ) = 0 ⟹ t = 1 , 3 t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0 \implies t = 1, 3 t 2 − 4 t + 3 = 0 ⟹ ( t − 1 ) ( t − 3 ) = 0 ⟹ t = 1 , 3
Interval Sign of v v v Motion 0 < t < 1 0 \lt t \lt 1 0 < t < 1 + + + Forward 1 < t < 3 1 \lt t \lt 3 1 < t < 3 − - − Backward 3 < t < 5 3 \lt t \lt 5 3 < t < 5 + + + Forward
D i s t a n c e = ∫ 0 1 v d t + ∣ ∫ 1 3 v d t ∣ + ∫ 3 5 v d t \mathrm{Distance} = \int_0^1 v\,dt + \left|\int_1^3 v\,dt\right| + \int_3^5 v\,dt Distance = ∫ 0 1 v d t + ∫ 1 3 v d t + ∫ 3 5 v d t = 4 3 + ∣ − 4 3 ∣ + 20 3 = 4 3 + 4 3 + 20 3 = 28 3 m = \frac{4}{3} + \left|-\frac{4}{3}\right| + \frac{20}{3} = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}\mathrm{ m} = 3 4 + − 3 4 + 3 20 = 3 4 + 3 4 + 3 20 = 3 28 m Improper Integrals Type 1: Infinite Limits ∫ a ∞ f ( x ) d x = lim b → ∞ ∫ a b f ( x ) d x \int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx ∫ a ∞ f ( x ) d x = b → ∞ lim ∫ a b f ( x ) d x Type 2: Discontinuous Integrand ∫ a b f ( x ) d x = lim t → a + ∫ t b f ( x ) d x \int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx ∫ a b f ( x ) d x = t → a + lim ∫ t b f ( x ) d x An improper integral converges if the limit exists (is finite) and diverges otherwise.
Example
Determine whether ∫ 1 ∞ 1 x 2 d x \displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx ∫ 1 ∞ x 2 1 d x
∫ 1 ∞ 1 x 2 d x = lim b → ∞ ∫ 1 b x − 2 d x = lim b → ∞ [ − 1 x ] 1 b = lim b → ∞ ( − 1 b + 1 ) = 1 \int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty}\int_1^b x^{-2}\,dx = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty}\left(-\frac{1}{b} + 1\right) = 1 ∫ 1 ∞ x 2 1 d x = b → ∞ lim ∫ 1 b x − 2 d x = b → ∞ lim [ − x 1 ] 1 b = b → ∞ lim ( − b 1 + 1 ) = 1 The integral converges to 1 1 1
Example
Determine whether ∫ 1 ∞ 1 x d x \displaystyle\int_1^{\infty} \frac{1}{x}\,dx ∫ 1 ∞ x 1 d x
∫ 1 ∞ 1 x d x = lim b → ∞ [ ln x ] 1 b = lim b → ∞ ln b = ∞ \int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty}[\ln x]_1^b = \lim_{b \to \infty}\ln b = \infty ∫ 1 ∞ x 1 d x = b → ∞ lim [ ln x ] 1 b = b → ∞ lim ln b = ∞ The integral diverges.
IB Exam-Style Questions Question 1 (Paper 1 style) Evaluate ∫ 0 π / 2 sin 2 x d x \displaystyle\int_0^{\pi/2} \sin^2 x\,dx ∫ 0 π /2 sin 2 x d x
Using the identity sin 2 x = 1 − cos 2 x 2 \sin^2 x = \dfrac{1 - \cos 2x}{2} sin 2 x = 2 1 − cos 2 x
∫ 0 π / 2 1 − cos 2 x 2 d x = 1 2 [ x − sin 2 x 2 ] 0 π / 2 = 1 2 ( π 2 − 0 ) = π 4 \int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{2}\left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4} ∫ 0 π /2 2 1 − cos 2 x d x = 2 1 [ x − 2 sin 2 x ] 0 π /2 = 2 1 ( 2 π − 0 ) = 4 π Question 2 (Paper 2 style) Let R R R y = x ( x − 2 ) y = x(x-2) y = x ( x − 2 ) x x x
(a) Find the area of R R R
Zeros: x = 0 x = 0 x = 0 x = 2 x = 2 x = 2
A r e a = ∫ 0 2 ∣ x ( x − 2 ) ∣ d x = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − x 3 3 ] 0 2 = 4 − 8 3 = 4 3 \mathrm{Area} = \int_0^2 |x(x-2)|\,dx = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3} Area = ∫ 0 2 ∣ x ( x − 2 ) ∣ d x = ∫ 0 2 ( 2 x − x 2 ) d x = [ x 2 − 3 x 3 ] 0 2 = 4 − 3 8 = 3 4 (b) Find the volume when R R R 360 ° 360\degree 360° x x x
V = π ∫ 0 2 [ x ( x − 2 ) ] 2 d x = π ∫ 0 2 x 2 ( x − 2 ) 2 d x V = \pi\int_0^2 [x(x-2)]^2\,dx = \pi\int_0^2 x^2(x-2)^2\,dx V = π ∫ 0 2 [ x ( x − 2 ) ] 2 d x = π ∫ 0 2 x 2 ( x − 2 ) 2 d x = π ∫ 0 2 ( x 4 − 4 x 3 + 4 x 2 ) d x = π [ x 5 5 − x 4 + 4 x 3 3 ] 0 2 = \pi\int_0^2 (x^4 - 4x^3 + 4x^2)\,dx = \pi\left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_0^2 = π ∫ 0 2 ( x 4 − 4 x 3 + 4 x 2 ) d x = π [ 5 x 5 − x 4 + 3 4 x 3 ] 0 2 = π ( 32 5 − 16 + 32 3 ) = π ⋅ 16 15 = 16 π 15 = \pi\left(\frac{32}{5} - 16 + \frac{32}{3}\right) = \pi \cdot \frac{16}{15} = \frac{16\pi}{15} = π ( 5 32 − 16 + 3 32 ) = π ⋅ 15 16 = 15 16 π Question 3 (Paper 2 style) Evaluate ∫ 2 x + 3 x 2 + 3 x + 2 d x \displaystyle\int \frac{2x + 3}{x^2 + 3x + 2}\,dx ∫ x 2 + 3 x + 2 2 x + 3 d x
Factorise the denominator: x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 ) x^2 + 3x + 2 = (x+1)(x+2) x 2 + 3 x + 2 = ( x + 1 ) ( x + 2 )
2 x + 3 ( x + 1 ) ( x + 2 ) = A x + 1 + B x + 2 \frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} ( x + 1 ) ( x + 2 ) 2 x + 3 = x + 1 A + x + 2 B 2 x + 3 = A ( x + 2 ) + B ( x + 1 ) 2x + 3 = A(x+2) + B(x+1) 2 x + 3 = A ( x + 2 ) + B ( x + 1 ) x = − 1 x = -1 x = − 1 1 = A ⟹ A = 1 1 = A \implies A = 1 1 = A ⟹ A = 1
x = − 2 x = -2 x = − 2 − 1 = − B ⟹ B = 1 -1 = -B \implies B = 1 − 1 = − B ⟹ B = 1
∫ 2 x + 3 x 2 + 3 x + 2 d x = ln ∥ x + 1 ∥ + ln ∥ x + 2 ∥ + C = ln ∥ ( x + 1 ) ( x + 2 ) ∥ + C \int \frac{2x+3}{x^2+3x+2}\,dx = \ln\|x+1\| + \ln\|x+2\| + C = \ln\|(x+1)(x+2)\| + C ∫ x 2 + 3 x + 2 2 x + 3 d x = ln ∥ x + 1∥ + ln ∥ x + 2∥ + C = ln ∥ ( x + 1 ) ( x + 2 ) ∥ + C Question 4 (Paper 1 style) Use integration by parts to evaluate ∫ 0 1 x e 2 x d x \displaystyle\int_0^1 xe^{2x}\,dx ∫ 0 1 x e 2 x d x
Let u = x$$dv = e^{2x}\,dx . Then du = dx$$v = \dfrac{1}{2}e^{2x} .
∫ 0 1 x e 2 x d x = [ 1 2 x e 2 x ] 0 1 − 1 2 ∫ 0 1 e 2 x d x = e 2 2 − 1 2 [ e 2 x 2 ] 0 1 \int_0^1 xe^{2x}\,dx = \left[\frac{1}{2}xe^{2x}\right]_0^1 - \frac{1}{2}\int_0^1 e^{2x}\,dx = \frac{e^2}{2} - \frac{1}{2}\left[\frac{e^{2x}}{2}\right]_0^1 ∫ 0 1 x e 2 x d x = [ 2 1 x e 2 x ] 0 1 − 2 1 ∫ 0 1 e 2 x d x = 2 e 2 − 2 1 [ 2 e 2 x ] 0 1 = e 2 2 − e 2 4 + 1 4 = e 2 4 + 1 4 = e 2 + 1 4 = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4} = 2 e 2 − 4 e 2 + 4 1 = 4 e 2 + 4 1 = 4 e 2 + 1 Question 5 (Paper 2 style) A particle moves in a straight line with acceleration a ( t ) = 6 t − 2 m / s 2 a(t) = 6t - 2\mathrm{ m/s}^2 a ( t ) = 6 t − 2 m/s 2 t = 0 t = 0 t = 0 4 m / s 4\mathrm{ m/s} 4 m/s 0 m 0\mathrm{ m} 0 m
(a) Find the velocity function.
V ( t ) = ∫ ( 6 t − 2 ) d t = 3 t 2 − 2 t + C V(t) = \int (6t-2)\,dt = 3t^2 - 2t + C V ( t ) = ∫ ( 6 t − 2 ) d t = 3 t 2 − 2 t + C v ( 0 ) = 4 ⟹ C = 4 v(0) = 4 \implies C = 4 v ( 0 ) = 4 ⟹ C = 4
V ( t ) = 3 t 2 − 2 t + 4 V(t) = 3t^2 - 2t + 4 V ( t ) = 3 t 2 − 2 t + 4 (b) Find the displacement function.
S ( t ) = ∫ ( 3 t 2 − 2 t + 4 ) d t = t 3 − t 2 + 4 t + D S(t) = \int (3t^2 - 2t + 4)\,dt = t^3 - t^2 + 4t + D S ( t ) = ∫ ( 3 t 2 − 2 t + 4 ) d t = t 3 − t 2 + 4 t + D s ( 0 ) = 0 ⟹ D = 0 s(0) = 0 \implies D = 0 s ( 0 ) = 0 ⟹ D = 0
S ( t ) = t 3 − t 2 + 4 t S(t) = t^3 - t^2 + 4t S ( t ) = t 3 − t 2 + 4 t (c) Find the total distance travelled in the first 3 seconds.
Check if v = 0 v = 0 v = 0 3 t 2 − 2 t + 4 = 0 3t^2 - 2t + 4 = 0 3 t 2 − 2 t + 4 = 0 = 4 − 48 < 0 = 4 - 48 \lt 0 = 4 − 48 < 0 v > 0 v \gt 0 v > 0
D i s t a n c e = ∫ 0 3 v d t = ∫ 0 3 ( 3 t 2 − 2 t + 4 ) d t = [ t 3 − t 2 + 4 t ] 0 3 = 27 − 9 + 12 = 30 m \mathrm{Distance} = \int_0^3 v\,dt = \int_0^3 (3t^2 - 2t + 4)\,dt = \left[t^3 - t^2 + 4t\right]_0^3 = 27 - 9 + 12 = 30\mathrm{ m} Distance = ∫ 0 3 v d t = ∫ 0 3 ( 3 t 2 − 2 t + 4 ) d t = [ t 3 − t 2 + 4 t ] 0 3 = 27 − 9 + 12 = 30 m Question 6 (Paper 2 style) The region bounded by y = e^x$$y = 1$$x = 0 And x = 2 x = 2 x = 2 x x x
V = π ∫ 0 2 [ ( e x ) 2 − 1 2 ] d x = π ∫ 0 2 ( e 2 x − 1 ) d x = π [ e 2 x 2 − x ] 0 2 V = \pi\int_0^2 [(e^x)^2 - 1^2]\,dx = \pi\int_0^2 (e^{2x} - 1)\,dx = \pi\left[\frac{e^{2x}}{2} - x\right]_0^2 V = π ∫ 0 2 [( e x ) 2 − 1 2 ] d x = π ∫ 0 2 ( e 2 x − 1 ) d x = π [ 2 e 2 x − x ] 0 2 = π ( e 4 2 − 2 − 1 2 + 0 ) = π ( e 4 2 − 5 2 ) = π ( e 4 − 5 ) 2 = \pi\left(\frac{e^4}{2} - 2 - \frac{1}{2} + 0\right) = \pi\left(\frac{e^4}{2} - \frac{5}{2}\right) = \frac{\pi(e^4 - 5)}{2} = π ( 2 e 4 − 2 − 2 1 + 0 ) = π ( 2 e 4 − 2 5 ) = 2 π ( e 4 − 5 ) Summary Table of Techniques Technique When to Use Key Idea Direct Simple power functions, standard forms Apply basic rules directly Substitution Composite functions, f ( g ( x ) ) g ′ ( x ) f(g(x))g'(x) f ( g ( x )) g ′ ( x ) Let u = g ( x ) u = g(x) u = g ( x ) By parts Product of different function types ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u Partial fractions Rational functions, factorisable denominator Decompose then integrate each term Trig substitution a 2 ± x 2 \sqrt{a^2 \pm x^2} a 2 ± x 2 x 2 − a 2 \sqrt{x^2 - a^2} x 2 − a 2 Replace with trig function
When facing an integral, first check if it is a standard form. If not, consider substitution (especially if you see a function and its derivative). If it is a product of different function Types, use integration by parts. If it is a rational function, consider partial fractions.
Additional Integration Techniques Integrals Involving Trigonometric Identities Many trigonometric integrals require using identities to simplify before integrating.
Powers of Sine and Cosine Odd power of sine : Factor out one sin x \sin x sin x sin 2 x = 1 − cos 2 x \sin^2 x = 1 - \cos^2 x sin 2 x = 1 − cos 2 x u = cos x u = \cos x u = cos x
Example
Evaluate ∫ sin 3 x d x \displaystyle\int \sin^3 x\,dx ∫ sin 3 x d x
∫ sin 2 x sin x d x = ∫ ( 1 − cos 2 x ) sin x d x \int \sin^2 x \sin x\,dx = \int (1 - \cos^2 x)\sin x\,dx ∫ sin 2 x sin x d x = ∫ ( 1 − cos 2 x ) sin x d x Let u = \cos x$$du = -\sin x\,dx :
= − ∫ ( 1 − u 2 ) d u = − u + u 3 3 + C = − cos x + cos 3 x 3 + C = -\int (1 - u^2)\,du = -u + \frac{u^3}{3} + C = -\cos x + \frac{\cos^3 x}{3} + C = − ∫ ( 1 − u 2 ) d u = − u + 3 u 3 + C = − cos x + 3 cos 3 x + C Even powers of sine or cosine : Use the half-angle formulas.
sin 2 x = 1 − cos 2 x 2 , cos 2 x = 1 + cos 2 x 2 \sin^2 x = \frac{1 - \cos 2x}{2}, \quad \cos^2 x = \frac{1 + \cos 2x}{2} sin 2 x = 2 1 − cos 2 x , cos 2 x = 2 1 + cos 2 x Example
Evaluate ∫ cos 4 x d x \displaystyle\int \cos^4 x\,dx ∫ cos 4 x d x
cos 4 x = ( cos 2 x ) 2 = ( 1 + cos 2 x 2 ) 2 = 1 + 2 cos 2 x + cos 2 2 x 4 \cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4} cos 4 x = ( cos 2 x ) 2 = ( 2 1 + cos 2 x ) 2 = 4 1 + 2 cos 2 x + cos 2 2 x = 1 + 2 cos 2 x + 1 + cos 4 x 2 4 = 3 + 4 cos 2 x + cos 4 x 8 = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8} = 4 1 + 2 cos 2 x + 2 1 + c o s 4 x = 8 3 + 4 cos 2 x + cos 4 x ∫ cos 4 x d x = 3 x 8 + sin 2 x 4 + sin 4 x 32 + C \int \cos^4 x\,dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C ∫ cos 4 x d x = 8 3 x + 4 sin 2 x + 32 sin 4 x + C Products of Sine and Cosine For ∫ sin m x cos n x d x \displaystyle\int \sin mx \cos nx\,dx ∫ sin m x cos n x d x
sin A cos B = 1 2 [ sin ( A + B ) + sin ( A − B ) ] \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] sin A cos B = 2 1 [ sin ( A + B ) + sin ( A − B )] sin A sin B = 1 2 [ cos ( A − B ) − cos ( A + B ) ] \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] sin A sin B = 2 1 [ cos ( A − B ) − cos ( A + B )] cos A cos B = 1 2 [ cos ( A − B ) + cos ( A + B ) ] \cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)] cos A cos B = 2 1 [ cos ( A − B ) + cos ( A + B )] Example
Evaluate ∫ sin 3 x cos 5 x d x \displaystyle\int \sin 3x \cos 5x\,dx ∫ sin 3 x cos 5 x d x
sin 3 x cos 5 x = 1 2 [ sin 8 x + sin ( − 2 x ) ] = 1 2 ( sin 8 x − sin 2 x ) \sin 3x \cos 5x = \frac{1}{2}[\sin 8x + \sin(-2x)] = \frac{1}{2}(\sin 8x - \sin 2x) sin 3 x cos 5 x = 2 1 [ sin 8 x + sin ( − 2 x )] = 2 1 ( sin 8 x − sin 2 x ) ∫ sin 3 x cos 5 x d x = 1 2 ( − cos 8 x 8 + cos 2 x 2 ) + C \int \sin 3x \cos 5x\,dx = \frac{1}{2}\left(-\frac{\cos 8x}{8} + \frac{\cos 2x}{2}\right) + C ∫ sin 3 x cos 5 x d x = 2 1 ( − 8 cos 8 x + 2 cos 2 x ) + C ∫ 1 a 2 + x 2 d x = 1 a arctan x a + C \int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C ∫ a 2 + x 2 1 d x = a 1 arctan a x + C ∫ 1 a 2 − x 2 d x = arcsin x a + C \int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C ∫ a 2 − x 2 1 d x = arcsin a x + C ∫ 1 x 2 − a 2 d x = 1 2 a ln ∣ x − a x + a ∣ + C \int \frac{1}{x^2 - a^2}\,dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C ∫ x 2 − a 2 1 d x = 2 a 1 ln x + a x − a + C Example
Evaluate ∫ 1 4 + 9 x 2 d x \displaystyle\int \frac{1}{4 + 9x^2}\,dx ∫ 4 + 9 x 2 1 d x
Rewrite as: ∫ 1 4 + ( 3 x ) 2 d x \displaystyle\int \frac{1}{4 + (3x)^2}\,dx ∫ 4 + ( 3 x ) 2 1 d x
Let u = 3x$$du = 3\,dx :
= 1 3 ∫ 1 4 + u 2 d u = 1 3 ⋅ 1 2 arctan u 2 + C = 1 6 arctan 3 x 2 + C = \frac{1}{3}\int \frac{1}{4 + u^2}\,du = \frac{1}{3} \cdot \frac{1}{2}\arctan\frac{u}{2} + C = \frac{1}{6}\arctan\frac{3x}{2} + C = 3 1 ∫ 4 + u 2 1 d u = 3 1 ⋅ 2 1 arctan 2 u + C = 6 1 arctan 2 3 x + C Integrals Involving e x e^x e x ln x \ln x ln x Integrals with e x e^x e x Use integration by parts when e x e^x e x
Integrals with ln x \ln x ln x Use integration by parts with u = ln x u = \ln x u = ln x d v = d x dv = dx d v = d x
∫ ln x d x = x ln x − x + C \int \ln x\,dx = x\ln x - x + C ∫ ln x d x = x ln x − x + C Example
Evaluate ∫ x 2 ln x d x \displaystyle\int x^2 \ln x\,dx ∫ x 2 ln x d x
Let u = \ln x$$dv = x^2\,dx . Then du = \dfrac{1}{x}\,dx$$v = \dfrac{x^3}{3} .
∫ x 2 ln x d x = x 3 3 ln x − ∫ x 3 3 ⋅ 1 x d x = x 3 3 ln x − 1 3 ∫ x 2 d x \int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx ∫ x 2 ln x d x = 3 x 3 ln x − ∫ 3 x 3 ⋅ x 1 d x = 3 x 3 ln x − 3 1 ∫ x 2 d x = x 3 3 ln x − x 3 9 + C = \frac{x^3}{3}\ln x - \frac{x^3}{9} + C = 3 x 3 ln x − 9 x 3 + C Arc Length The arc length of a curve y = f ( x ) y = f(x) y = f ( x ) x = a x = a x = a x = b x = b x = b
L = \int_a^b \sqrt{1 + \left(\frac`\{dy}``\{dx}`\right)^2}\,dx For a parametric curve ( x ( t ) , y ( t ) ) (x(t), y(t)) ( x ( t ) , y ( t )) t = t 1 t = t_1 t = t 1 t = t 2 t = t_2 t = t 2
L = \int_{t_1}^{t_2} \sqrt{\left(\frac`\{dx}``\{dt}`\right)^2 + \left(\frac`\{dy}``\{dt}`\right)^2}\,dt Example
Find the arc length of y = x 3 6 + 1 2 x y = \dfrac{x^3}{6} + \dfrac{1}{2x} y = 6 x 3 + 2 x 1 x = 1 x = 1 x = 1 x = 3 x = 3 x = 3
\frac`\{dy}``\{dx}` = \frac{x^2}{2} - \frac{1}{2x^2} \left(\frac`\{dy}``\{dx}`\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4} 1 + \left(\frac`\{dy}``\{dx}`\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2 L = ∫ 1 3 ( x 2 2 + 1 2 x 2 ) d x = [ x 3 6 − 1 2 x ] 1 3 L = \int_1^3 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)\,dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^3 L = ∫ 1 3 ( 2 x 2 + 2 x 2 1 ) d x = [ 6 x 3 − 2 x 1 ] 1 3 = ( 27 6 − 1 6 ) − ( 1 6 − 1 2 ) = 26 6 + 1 3 = 14 3 = \left(\frac{27}{6} - \frac{1}{6}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) = \frac{26}{6} + \frac{1}{3} = \frac{14}{3} = ( 6 27 − 6 1 ) − ( 6 1 − 2 1 ) = 6 26 + 3 1 = 3 14 Mean Value of a Function The mean value of a function f ( x ) f(x) f ( x ) [ a , b ] [a, b] [ a , b ]
f ˉ = 1 b − a ∫ a b f ( x ) d x \bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx f ˉ = b − a 1 ∫ a b f ( x ) d x Example
Find the mean value of f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 [ 0 , 3 ] [0, 3] [ 0 , 3 ]
f ˉ = 1 3 ∫ 0 3 x 2 d x = 1 3 [ x 3 3 ] 0 3 = 1 3 × 9 = 3 \bar{f} = \frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \times 9 = 3 f ˉ = 3 1 ∫ 0 3 x 2 d x = 3 1 [ 3 x 3 ] 0 3 = 3 1 × 9 = 3 Additional Exam-Style Questions Question 7 (Paper 2 style) Evaluate ∫ 0 π / 4 x cos 2 x d x \displaystyle\int_0^{\pi/4} x\cos 2x\,dx ∫ 0 π /4 x cos 2 x d x
Let u = x$$dv = \cos 2x\,dx . Then du = dx$$v = \dfrac{\sin 2x}{2} .
∫ 0 π / 4 x cos 2 x d x = [ x sin 2 x 2 ] 0 π / 4 − 1 2 ∫ 0 π / 4 sin 2 x d x \int_0^{\pi/4} x\cos 2x\,dx = \left[\frac{x\sin 2x}{2}\right]_0^{\pi/4} - \frac{1}{2}\int_0^{\pi/4}\sin 2x\,dx ∫ 0 π /4 x cos 2 x d x = [ 2 x sin 2 x ] 0 π /4 − 2 1 ∫ 0 π /4 sin 2 x d x = ( π / 4 × 1 2 − 0 ) − 1 2 [ − cos 2 x 2 ] 0 π / 4 = \left(\frac{\pi/4 \times 1}{2} - 0\right) - \frac{1}{2}\left[-\frac{\cos 2x}{2}\right]_0^{\pi/4} = ( 2 π /4 × 1 − 0 ) − 2 1 [ − 2 cos 2 x ] 0 π /4 = π 8 − 1 2 ( − 0 2 + 1 2 ) = π 8 − 1 4 = \frac{\pi}{8} - \frac{1}{2}\left(-\frac{0}{2} + \frac{1}{2}\right) = \frac{\pi}{8} - \frac{1}{4} = 8 π − 2 1 ( − 2 0 + 2 1 ) = 8 π − 4 1 Question 8 (Paper 2 style) Find the area enclosed by the curves y = x 3 y = x^3 y = x 3 y = x y = \sqrt{x} y = x
Intersection: x 3 = x ⟹ x 6 = x ⟹ x ( x 5 − 1 ) = 0 ⟹ x = 0 x^3 = \sqrt{x} \implies x^6 = x \implies x(x^5 - 1) = 0 \implies x = 0 x 3 = x ⟹ x 6 = x ⟹ x ( x 5 − 1 ) = 0 ⟹ x = 0 x = 1 x = 1 x = 1
For 0 ≤ x ≤ 1 0 \le x \le 1 0 ≤ x ≤ 1 x ≥ x 3 \sqrt{x} \ge x^3 x ≥ x 3
A r e a = ∫ 0 1 ( x − x 3 ) d x = [ 2 x 3 / 2 3 − x 4 4 ] 0 1 = 2 3 − 1 4 = 5 12 \mathrm{Area} = \int_0^1 (\sqrt{x} - x^3)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^4}{4}\right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12} Area = ∫ 0 1 ( x − x 3 ) d x = [ 3 2 x 3/2 − 4 x 4 ] 0 1 = 3 2 − 4 1 = 12 5 Question 9 (Paper 1 style) Evaluate ∫ x x 2 + 2 x + 5 d x \displaystyle\int \frac{x}{x^2 + 2x + 5}\,dx ∫ x 2 + 2 x + 5 x d x
Complete the square: x 2 + 2 x + 5 = ( x + 1 ) 2 + 4 x^2 + 2x + 5 = (x+1)^2 + 4 x 2 + 2 x + 5 = ( x + 1 ) 2 + 4
Let u = x^2 + 2x + 5$$du = (2x + 2)\,dx .
We need to split: x x 2 + 2 x + 5 = ( 2 x + 2 ) / 2 − 1 x 2 + 2 x + 5 = 1 2 ⋅ 2 x + 2 x 2 + 2 x + 5 − 1 x 2 + 2 x + 5 \dfrac{x}{x^2+2x+5} = \dfrac{(2x+2)/2 - 1}{x^2+2x+5} = \dfrac{1}{2} \cdot \dfrac{2x+2}{x^2+2x+5} - \dfrac{1}{x^2+2x+5} x 2 + 2 x + 5 x = x 2 + 2 x + 5 ( 2 x + 2 ) /2 − 1 = 2 1 ⋅ x 2 + 2 x + 5 2 x + 2 − x 2 + 2 x + 5 1
∫ x x 2 + 2 x + 5 d x = 1 2 ln ( x 2 + 2 x + 5 ) − ∫ 1 ( x + 1 ) 2 + 4 d x \int \frac{x}{x^2+2x+5}\,dx = \frac{1}{2}\ln(x^2+2x+5) - \int \frac{1}{(x+1)^2 + 4}\,dx ∫ x 2 + 2 x + 5 x d x = 2 1 ln ( x 2 + 2 x + 5 ) − ∫ ( x + 1 ) 2 + 4 1 d x = 1 2 ln ( x 2 + 2 x + 5 ) − 1 2 arctan ( x + 1 2 ) + C = \frac{1}{2}\ln(x^2+2x+5) - \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C = 2 1 ln ( x 2 + 2 x + 5 ) − 2 1 arctan ( 2 x + 1 ) + C Question 10 (Paper 2 style) The region bounded by y = \ln x$$y = 0 And x = e x = e x = e 360 ° 360\degree 360° x x x
V = π ∫ 1 e ( ln x ) 2 d x V = \pi\int_1^e (\ln x)^2\,dx V = π ∫ 1 e ( ln x ) 2 d x Using integration by parts with u = (\ln x)^2$$dv = dx :
= π [ x ( ln x ) 2 ] 1 e − 2 π ∫ 1 e ln x d x = \pi\left[x(\ln x)^2\right]_1^e - 2\pi\int_1^e \ln x\,dx = π [ x ( ln x ) 2 ] 1 e − 2 π ∫ 1 e ln x d x = π ( e − 0 ) − 2 π [ x ln x − x ] 1 e = π e − 2 π ( e − e + 1 ) = π e − 2 π = π ( e − 2 ) = \pi(e - 0) - 2\pi[x\ln x - x]_1^e = \pi e - 2\pi(e - e + 1) = \pi e - 2\pi = \pi(e - 2) = π ( e − 0 ) − 2 π [ x ln x − x ] 1 e = π e − 2 π ( e − e + 1 ) = π e − 2 π = π ( e − 2 ) For the A-Level treatment of this topic, see Integration .
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Common Pitfalls Forgetting the chain rule when integrating composite functions — look for an inner function and its derivative.
Confusing definite and indefinite integrals — definite integrals give a numerical value; indefinite integrals give a family of functions.
Incorrectly applying limits of integration after a substitution — remember to change the limits to the new variable.
Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.
Cancelling terms instead of factors — a b + a c a \frac{ab + ac}{a} a ab + a c b + c b + c b + c b c bc b c
Summary Indefinite integral always includes + C +C + C Area between curves: ∫ ∣ f ( x ) − g ( x ) ∣ d x \int |f(x) - g(x)|\,dx ∫ ∣ f ( x ) − g ( x ) ∣ d x Volume of revolution: V = π ∫ y 2 d x V = \pi\int y^2\,dx V = π ∫ y 2 d x Techniques: substitution, integration by parts (LIATE), partial fractions Worked Examples Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.