Resources for Integration by Substitution
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Video Tutorials
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Integration by Substitution Theory
![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)
NSW Y12 Maths - Extension 1 Calculus Integration by Substitution
Videos
Videos relating to Integration by Substitution.
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Integration by Substitution - Video - Lesson
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Integration by Substitution - Video - Exercises
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Integration by Substitution - Video - Integration by Substitution Introduction
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Integration by Substitution - Video - Integration by substitution worked examples part 1
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Integration by Substitution - Video - Integration by substitution worked examples part 2
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