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With Worked Solution![To integrate \(I=\displaystyle \int_1^3 x(x-1)^4 d x\) it would he tedious to expand and then integrate. \\ Instead, if we let \(u=x-1\) and also differentiate \(u=x-1\) to get \(\dfrac{d u}{d x}=1\)\\ Then I becomes \(\displaystyle \int_{x=1}^{x=3}(u+1) u^4 d u \quad(\) since \(d u=d x)\).\\ The limits also must be changed\\ $\begin{aligned} \therefore I & =\displaystyle \int_0^2 u^5+u^4 d u \\ & =\left[\frac{1}{6} u^6+\frac{1}{5} u^5\right]_0^2 \\ & =\frac{1}{6} \times 2^6+\frac{1}{5} \times 2^5 \\ & =\frac{256}{15} . \end{aligned}$\\ \textbf{Example 1} %(21412} Using the substitution \(u = {x^2} + 2\), evaluate \(\displaystyle \int\limits_0^{\sqrt 2 } {x\sqrt {{x^2} + 2} \,\,dx} \)\\ \begin{multicols}{2} \textbf{Example 1 solution}\\ \(I=\int^{\sqrt{2}}_{0}x\sqrt{x^2+2}dx\)\\ $\begin{aligned} x&=\sqrt{2}\rightarrow u=4\\ x&= 0 \rightarrow u=2 \end{aligned}$\\ $\begin{aligned} u&=x^2+2\\ \frac{du}{dx}&=2x\\ \frac{1}{2}du&=xdx \end{aligned}$\\ \columnbreak \textbf{Example cont}\\ $\begin{aligned} I&=\frac{1}{2}\displaystyle \int^{4}_{2}\sqrt{u}\,du\\ &=\frac{1}{2}\displaystyle \int^{4}_{2}u^{\frac{1}{2}}\,du\\ &=\frac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\right]^{4}_{2}\\ &=\frac{1}{3}[4^{\frac{3}{2}}-2^{\frac{3}{2}}]\\ &=\frac{1}{3}[8-2\sqrt{2}]\\ &=\frac{2}{3}[4-\sqrt{2}] \end{aligned}$\\ \end{multicols}](/media/43kdeoqs/integration-by-substitution.png)
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The method of substitution involves transforming an integral with respect to one variable, say \(x\), into an integral with respect to a related variable, say \(u\).
Let \(y = \int {f(x)\,dx} \) then \(\dfrac{{dy}}{{dx}} = f(x)\)
If \(u\) is a function of \(x\), then, by the chain rule \(\dfrac{{dy}}{{du}} = \dfrac{{dy}}{{dx}} \times \dfrac{{dx}}{{du}} \to f(x) \times \dfrac{{dx}}{{du}}\)
Integrating with respect to \(u\) gives \(y = \int {f(x)\dfrac{{dx}}{{du}}} \,\,du\)
This gives the result used in the method known as substitution or change of variable.
For example: Find \(\int {x{{\left( {2x + 1} \right)}^5}} dx\) using the substitution \(u = 2x + 1\)
\(I = \int {x{{\left( {2x + 1} \right)}^5}} dx\) Let \(u = 2x + 1 \to x = \dfrac{1}{2}\left( {u - 1} \right) \to \dfrac{{dx}}{{du}} = \dfrac{1}{2}\).
\(\begin{array}{l}I = \int {x{{(2x + 1)}^5}} \dfrac{{dx}}{{du}}\,du\\\,\,\,\,\, = \int {\dfrac{1}{2}} \left( {u - 1} \right){u^5} \times \dfrac{1}{2}\,du\\\,\,\,\, = \dfrac{1}{4}\int {\left( {{u^5} - {u^6}} \right)\,du} \,\\\,\,\,\, = \dfrac{1}{4}\left( {\dfrac{{{u^6}}}{6} - \dfrac{{{u^7}}}{7}} \right) = \dfrac{1}{4}\left\{ {\dfrac{{{{\left( {2x + 1} \right)}^6}}}{6} - \dfrac{{{{\left( {2x + 1} \right)}^7}}}{7}} \right\} = \dfrac{1}{{168}}\left( {12x - 1} \right){\left( {2x + 1} \right)^6} + c\\\,\end{array}\)
The Definite Integral
With the definite integral it is usual to change the limits at the same time as the variable is changed, rather than to integrate and revert to the original variable and the corresponding limits.
For example: Evaluate \(\int\limits_0^{\log 4} {\dfrac{{{e^{2x}}}}{{{e^x} + 2}}} \,\,dx\,\,\,\)
Let \(u = {e^x} + 2 \to {e^x} = u - 2\)
When \(x = \log 4\,\, \to u = {e^{\log 4}} + 2 \to u = 6\)
\(x = 0\,\, \to u = {e^o} + 2 \to u = 3\)
\(u = {e^x} + 2 \to \dfrac{{du}}{{dx}} = {e^x} \to \dfrac{{dx}}{{du}} = \dfrac{1}{{{e^x}}}\)
\(\begin{array}{l}I = \int\limits_0^{\log 4} {\dfrac{{{e^{2x}}}}{{{e^x} + 2}}} \,\dfrac{{dx}}{{du}}\,du\\\,\,\,\,\, = \int\limits_3^6 {\dfrac{{{e^{2x}}}}{u}} \times \dfrac{1}{{{e^x}}}\,du = \int\limits_3^6 {\dfrac{{\left( {u - 2} \right)}}{u}} \,\,du\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int\limits_3^6 {1 - \dfrac{2}{u}} \,\,du\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {u - 2\log u} \right]_3^6 = \left( {6 - 2\log 6} \right) - \left( {3 - 2\log 3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3 - 2\log 2\end{array}\)