Mathematics

Integrate $$\sqrt { 1-2x } dx$$


SOLUTION
$$\displaystyle I=\int \sqrt{1-2x}dx$$
Substitute $$\sqrt{1-2x}=t\Rightarrow 1-2x=t^2\Rightarrow -2dx=2tdt\Rightarrow -dx=tdt$$
$$\displaystyle I=\int t(-tdt)=-\int t^2dt=-\dfrac{t^3}{3}+C$$
Undo $$t=\sqrt{1-2x}$$
$$\displaystyle I=-\dfrac{(\sqrt{1-2x})^3}{3}+C$$
$$\displaystyle I=-\dfrac{(1-2x)^{3/2}}{3}+C$$
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Subjective Medium Published on 17th 09, 2020
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