Evaluate the following integrals : $\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x$
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Answer
Let $I=\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x$ We use the property, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ Hence in $I$, we replace $x$ by $1+3-x$. $ \begin{aligned} \therefore I & =\int_1^3 \frac{\sqrt[3]{1+3-x+5}}{\sqrt[3]{1+3-x+5}+\sqrt[3]{9-1-3+x}} d x \\ & =\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{x+5}} d x \end{aligned} $ Adding (1) and (2), we get $ \begin{aligned} 2 I & =\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x+\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{x+5}} d x \\ & =\int_1^3 \frac{\sqrt[3]{x+5}+\sqrt[3]{9-x}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x \\ & =\int_1^3 1 d x=[x]_1^3 \\ & =3-1=2 \\ \therefore & I=1 \end{aligned} $ Hence, $\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x=1$.
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