Indefinite Integration
The reverse of differentiation · Power rule · Constant of integration · §15.1
The power rule for integration
\(\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C\quad(n\neq-1)\)
Add 1 to the power, divide by the new power, add constant C. Integration is the reverse of differentiation.
Always add +C for indefinite integrals. This represents any constant that differentiated to zero. Forgetting C loses marks.
Exception: \(n=-1\):
\(\int \dfrac{1}{x}\,dx = \ln|x|+C\)
\(\int \dfrac{1}{x}\,dx = \ln|x|+C\)
Standard integrals
\(\int e^x\,dx\)\(e^x+C\)
\(\int e^{ax}\,dx\)\(\dfrac{1}{a}e^{ax}+C\)
\(\int \sin x\,dx\)\(-\cos x+C\)
\(\int \cos x\,dx\)\(\sin x+C\)
\(\int \sec^2 x\,dx\)\(\tan x+C\)
\(\int (ax+b)^n\,dx\)\(\dfrac{(ax+b)^{n+1}}{a(n+1)}+C\)
Find \(\int 3x^2 - 5x + \dfrac{2}{x^3}\,dx\)
Rewrite
\(= \int 3x^2-5x+2x^{-3}\,dx\)
Integrate
\(= x^3 - \dfrac{5x^2}{2} + \dfrac{2x^{-2}}{-2}+C\)
✓
\(= x^3-\dfrac{5x^2}{2}-\dfrac{1}{x^2}+C\)
Given \(\dfrac{dy}{dx}=3x^2-4\) and y=5 when x=1, find y.
Integrate
\(y = x^3-4x+C\)
Find C
Sub x=1, y=5: \(5=1-4+C\Rightarrow C=8\)
✓
\(y=x^3-4x+8\)
Definite Integration
Evaluating between limits · Area under a curve · §15.2
Definite integral
\(\int_a^b f(x)\,dx = [F(x)]_a^b = F(b)-F(a)\)
No constant C needed — it cancels. The result is a number, not a function.
Order matters: \(\int_a^b = -\int_b^a\). Swapping limits negates the result.
Negative areas: If the curve is below the x-axis between a and b, the definite integral is negative. For area, take the absolute value.
Evaluate \(\displaystyle\int_1^3(x^2-2x+1)\,dx\)
Integrate
\(\left[\dfrac{x^3}{3}-x^2+x\right]_1^3\)
Upper (x=3)
\(9-9+3=3\)
Lower (x=1)
\(\tfrac{1}{3}-1+1=\tfrac{1}{3}\)
✓
\(3-\tfrac{1}{3}=\dfrac{8}{3}\)
Area Under & Between Curves
Area = ∫y dx · Area between two curves · Trapezium rule · §15.3
Area formulas
Area under curve (above x-axis):
\(A = \displaystyle\int_a^b y\,dx\)
Area between two curves:
\(A = \displaystyle\int_a^b (y_{\text{top}}-y_{\text{bot}})\,dx\)
Trapezium rule (approximation):
\(A \approx \tfrac{1}{2}h[y_0+2(y_1+\cdots+y_{n-1})+y_n]\)
Area enclosed by \(y=x^2\) and \(y=x+2\)
Intersect
\(x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow(x-2)(x+1)=0\)
\(x=-1\) and \(x=2\)
\(x=-1\) and \(x=2\)
Area
\(A=\displaystyle\int_{-1}^{2}(x+2-x^2)\,dx\)
Integrate
\(=\left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2}\)
Evaluate
At x=2: \(2+4-\tfrac{8}{3}=\tfrac{10}{3}\). At x=-1: \(\tfrac{1}{2}-2+\tfrac{1}{3}=-\tfrac{7}{6}\)
✓
\(A=\tfrac{10}{3}+\tfrac{7}{6}=\dfrac{27}{6}=\dfrac{9}{2}\)
Interactive — Area under \(y=ax^n\) from 0 to b
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Practice Questions
3 Easy · 3 Medium · 3 Hard — Integration
Easy 1
[2] Find \(\displaystyle\int(3x^2-4x+1)\,dx\).
SOLUTION
Integrate
\(\boldsymbol{x^3-2x^2+x+c}\)
Easy 2
[3] Evaluate \(\displaystyle\int_1^3(2x-1)\,dx\).
SOLUTION
Integrate
\([x^2-x]_1^3=(9-3)-(1-1)=6-0=\boldsymbol{6}\)
Easy 3
[3] Find \(\displaystyle\int(2x+1)^4\,dx\).
SOLUTION — reverse chain rule
Integrate
\(\dfrac{(2x+1)^5}{5\times2}+c=\boldsymbol{\dfrac{(2x+1)^5}{10}+c}\)
Medium 1
[4] Find the area enclosed between \(y=x^2\) and \(y=4x-x^2\).
SOLUTION
Intersections
\(x^2=4x-x^2\Rightarrow2x^2-4x=0\Rightarrow x=0\) or \(x=2\)
Area
\(\displaystyle\int_0^2(4x-x^2-x^2)\,dx=\int_0^2(4x-2x^2)\,dx=[2x^2-\tfrac{2x^3}{3}]_0^2=(8-\tfrac{16}{3})=\boldsymbol{\tfrac{8}{3}}\)
Medium 2
[4] Find \(\displaystyle\int_0^{\pi/2}3\sin(2x+\tfrac{\pi}{4})\,dx\).
SOLUTION
Integrate
\(\left[-\dfrac{3}{2}\cos(2x+\tfrac{\pi}{4})\right]_0^{\pi/2}\)
Evaluate
\(=-\tfrac32\cos(\pi+\tfrac\pi4)+\tfrac32\cos(\tfrac\pi4)=-\tfrac32(-\frac{\sqrt2}{2})+\tfrac32(\frac{\sqrt2}{2})=\tfrac{3\sqrt2}{4}+\tfrac{3\sqrt2}{4}=\boldsymbol{\dfrac{3\sqrt2}{2}}\)
Medium 3
[4] Find \(\displaystyle\int\frac{2}{3x-1}\,dx\) and \(\displaystyle\int3e^{2x-1}\,dx\).
SOLUTION
Log form
\(\displaystyle\int\dfrac{2}{3x-1}\,dx=\dfrac{2}{3}\ln|3x-1|+c\)
Exponential
\(\displaystyle\int3e^{2x-1}\,dx=\dfrac{3}{2}e^{2x-1}+c\)
Hard 1
[5] Find the area bounded by \(y=\frac{4}{x}\), the x-axis, and lines \(x=1\) and \(x=e^2\).
SOLUTION
Integrate
\(\displaystyle\int_1^{e^2}\dfrac{4}{x}\,dx=4[\ln x]_1^{e^2}=4(\ln e^2-\ln1)=4(2-0)=\boldsymbol{8}\)
Hard 2
[6] Find the volume generated when the region bounded by \(y=\sqrt{x}\), the x-axis, and \(x=4\) is rotated \(2\pi\) radians about the x-axis.
SOLUTION — volume of revolution
Formula
\(V=\pi\displaystyle\int_0^4 y^2\,dx=\pi\int_0^4 x\,dx\)
Evaluate
\(=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\times8=\boldsymbol{8\pi}\) units³
Hard 3
[6] A curve passes through \((0,3)\) with gradient \(\frac{dy}{dx}=6x^2-4x\). Find the equation of the curve and the area between the curve and the x-axis between its roots.
SOLUTION
Integrate
\(y=2x^3-2x^2+c\). At \((0,3)\): \(c=3\). Curve: \(\boldsymbol{y=2x^3-2x^2+3}\)
Roots
\(2x^3-2x^2+3=0\). By inspection/numerical: one real root \(\approx-0.88\). No simple roots — area between roots would need numerical integration.
Past Year Paper Questions
Cambridge 0606 — Integration
0606
[11]
(a) Show that \(\displaystyle\int_0^{\pi/4}\tan^2x\,dx=1-\dfrac{\pi}{4}\). [Hint: \(\tan^2x=\sec^2x-1\)] [4]
(b) Find the area enclosed by \(y=e^{2x}-1\) and \(y=3x\). [5]
(c) Evaluate \(\displaystyle\int_1^4\dfrac{x^2+2}{\sqrt{x}}\,dx\). [2]
(b) Find the area enclosed by \(y=e^{2x}-1\) and \(y=3x\). [5]
(c) Evaluate \(\displaystyle\int_1^4\dfrac{x^2+2}{\sqrt{x}}\,dx\). [2]
FULL SOLUTION
(a)
\(\displaystyle\int_0^{\pi/4}(\sec^2x-1)\,dx=[\tan x-x]_0^{\pi/4}=(1-\tfrac\pi4)-(0)=1-\tfrac\pi4\) ✓
(b) Intersect
Numerically: \(e^{2x}-1=3x\) at \(x=0\) and \(x\approx1.256\)
(b) Area
\(\displaystyle\int_0^{1.256}(3x-e^{2x}+1)\,dx=[\frac{3x^2}{2}-\frac{e^{2x}}{2}+x]_0^{1.256}\approx\boldsymbol{0.430}\)
(c)
\(\displaystyle\int_1^4(x^{3/2}+2x^{-1/2})\,dx=[\tfrac25x^{5/2}+4x^{1/2}]_1^4=({\tfrac25\times32+4\times2})-(\tfrac25+4)=(\tfrac{64}{5}+8)-\tfrac{22}{5}=\tfrac{42}{5}+8=\boldsymbol{\tfrac{82}{5}}\)
0606 Style
Volume of Revolution
[8]
The region \(R\) is bounded by \(y = \sqrt{x+1}\), the \(x\)-axis, \(x=0\) and \(x=3\).
(a) Find the area of \(R\). [3]
(b) Find the volume when \(R\) is rotated \(2\pi\) radians about the \(x\)-axis. [3]
(c) Find the volume when \(R\) is rotated \(2\pi\) radians about the \(y\)-axis. (Set up only.) [2]
(a) Find the area of \(R\). [3]
(b) Find the volume when \(R\) is rotated \(2\pi\) radians about the \(x\)-axis. [3]
(c) Find the volume when \(R\) is rotated \(2\pi\) radians about the \(y\)-axis. (Set up only.) [2]
FULL WORKED SOLUTION
(a)
\(\int_0^3\sqrt{x+1}\,dx=\left[\frac{2}{3}(x+1)^{3/2}\right]_0^3=\frac{2}{3}(8-1)=\boldsymbol{\frac{14}{3}}\)
(b)
\(V=\pi\int_0^3(x+1)dx=\pi\left[\frac{x^2}{2}+x\right]_0^3=\pi(\frac{9}{2}+3)=\boldsymbol{\frac{15\pi}{2}}\)
(c)
\(V=\pi\int_1^2 x^2\,dy\) where \(x=y^2-1\). \(V=\pi\int_1^2(y^2-1)^2\,dy\)
0606 Style
Integration by Parts (intro)
[7]
(a) Find \(\displaystyle\int x e^x\,dx\) using integration by parts. [4]
(b) Hence evaluate \(\displaystyle\int_0^2 xe^x\,dx\). [3]
(b) Hence evaluate \(\displaystyle\int_0^2 xe^x\,dx\). [3]
FULL WORKED SOLUTION
(a)
Let \(u=x\), \(dv=e^x dx\). Then \(du=dx\), \(v=e^x\).
\(\int xe^x dx=xe^x-\int e^x dx=\boldsymbol{xe^x-e^x+c=(x-1)e^x+c}\)
\(\int xe^x dx=xe^x-\int e^x dx=\boldsymbol{xe^x-e^x+c=(x-1)e^x+c}\)
(b)
\(\left[(x-1)e^x\right]_0^2=(1)e^2-(-1)e^0=e^2+1\approx\boldsymbol{8.39}\)