Evaluate _1^3 [3] x+5 [3] x+5 + [3] 9-x d x - Vidyadip

Question

Evaluate $\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} d x$

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Answer

Let $I =\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} \cdot d x$
Here we use the property,
$ \int_{ a }^{ b } f (x) \cdot dx =\int_{ a }^{ b } f ( a + b -x) \cdot dx$
$\therefore \text { Replacing } x \text { in (i) by }(1+3- x )$
$=\int_1^3 \frac{\sqrt[3]{(1+3-x)+5}}{\sqrt[3]{(1+3-x)+5}+\sqrt[3]{9-(1+3-x)}} \cdot d x$
$\therefore I =\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{x+5}} \cdot d x \quad \ldots . . \text { (ii) } $
Adding (i) and (ii), we get
$ \text {2I }=\int_1^3 \frac{\sqrt[3]{x+5}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} \cdot d x+\int_1^3 \frac{\sqrt[3]{9-x}}{\sqrt[3]{9-x}+\sqrt[3]{5+x}} \cdot d x$
$=\int_1^3 \frac{\sqrt[3]{x+5}+\sqrt[3]{9-x}}{\sqrt[3]{x+5}+\sqrt[3]{9-x}} \cdot d x$
$=\int_1^3 1 \cdot d x=[x]_1^3$
$\therefore 2 I =3-1=2$
$\therefore I =1 $

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