# Definite Integration by Substitution

## Definite Integration by Substitution

Definite Integration by Substitution requires converting definite integration’s upper and lower limits.
$$\large \displaystyle \int_{x=a}^{x=b}{f(x)}dx = \int_{u=c}^{u=d}{f(u)}du$$

# Practice Questions of Definite Integration by Substitution

### Question 1

Find $\displaystyle \int_{-1}^{0}{x(1+x)^{10}}dx$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{Let } u &= 1+x \\ \frac{du}{dx} &= 1 \\ du &= dx \\ x &= u-1 &\color{red} u = 1+x \\ u &= 1+0 = 1 &\color{red} \text{for } x=0 \\ u &= 1-1 = 0 &\color{red} \text{for } x=-1 \\ \int_{-1}^{0}{x(1+x)^{10}}dx &= \int_{0}^{1}{(u-1)u^{10}}du \\ &= \int_{0}^{1}{(u^{11}-u^{10})}du \\ &= \Big[\frac{1}{12}u^{12}-\frac{1}{11}u^{11}\Big]_{0}^{1} \\ &= \frac{1}{12}-\frac{1}{11} \\ &= -\frac{1}{132} \end{aligned}

### Question 2

Find $\displaystyle \int_{0}^{1}{x \sqrt{1-x^2}}dx$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{Let } u &= 1-x^2 \\ \frac{du}{dx} &= -2x \\ \frac{du}{-2x} &= dx \\ u &= 1-1^2 &\color{red} \text{for } x=1 \\ u &= 1-0^2 &\color{red} \text{for } x=0 \\ \int_{0}^{1}{x \sqrt{1-x^2}}dx &= \int_{1}^{0}{x \sqrt{u}} \frac{du}{-2x} \\ &= -\frac{1}{2} \int_{1}^{0}{\sqrt{u}}du \\ &= \frac{1}{2} \int_{0}^{1}{\sqrt{u}}du &\color{red}-\int_{a}^{b} = \int_{b}^{a} \\ &= \frac{1}{2} \int_{0}^{1}{u^{\frac{1}{2}}}du \\ &= \frac{1}{2} \Bigg[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\Bigg]_{0}^{1} \\ &= \frac{1}{3}\Big[\sqrt{u^3}\Big]_{0}^{1} \\ &= \frac{1}{3}\Big[\sqrt{1^3}-\sqrt{0^3}\Big] \\ &= \frac{1}{3} \end{aligned}

### Question 3

Find $\displaystyle \int_{3}^{18}{\frac{x}{\sqrt{x-2}}}dx$.

\begin{aligned} \require{AMSsymbols} \displaystyle \text{Let } u &= \sqrt{x-2} \\ u^2 &= x-2 \\ x &= u^2 + 2 \\ \frac{dx}{du} &= 2u \\ dx &= 2udu \\ u &= \sqrt{18-2} = 4 &\color{red} \text{for } x=18 \\ u &= \sqrt{3-2} = 1 &\color{red} \text{for } x=3 \\ \int_{3}^{18}{\frac{x}{\sqrt{x-2}}}dx &= \int_{1}^{4}{\frac{u^2+2}{u}2u}du \\ &= \int_{1}^{4}{\big(2u^2 + 4\big)}du \\ &= \Big[\frac{2}{3}u^3 + 4u\Big]_{1}^{4} \\ &= \Big[\frac{2}{3} \times 4^3 + 4 \times 3\Big]-\Big[\frac{2}{3} \times 1^3 + 4 \times 1\Big] \\ &= 54 \end{aligned}

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