Mathematics

IF $$f(x)=x^{2}$$ for $$0 \le x \le 1,\sqrt {x}$$ for $$1 \le x \le 2$$ then $$\int _{0}^{2}f(x)dx=$$


ANSWER

$$\dfrac {4\sqrt {2}-1}{3}$$


SOLUTION
$$\int_0^2 {f\left( x \right) = } \int_0^1 {f\left( x \right)dx + \int_1^2 {f\left( x \right)dx} } $$
$$ = \int_0^1 {{x^2}dx + \int_1^2 {\sqrt x dx} } $$
$$ = \left[ {\cfrac{{{x^3}}}{3}} \right]_0^1 + \left[ {\cfrac{{{x^{\cfrac{3}{2}}}}}{{\cfrac{3}{2}}}} \right]_1^2$$
$$ = \cfrac{1}{3}\left( 1 \right) + \cfrac{2}{3}\left( {2\sqrt 2  - 1} \right)$$
$$ = \cfrac{{4\sqrt 2  - 2 + 1}}{3}$$
$$ = \cfrac{{4\sqrt 2  - 1}}{3}$$
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Single Correct Medium Published on 17th 09, 2020
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