Mathematics

$$\displaystyle \int _{0}^{1}(\sqrt{x})dx+\displaystyle \int _{0}^{4/3}(\sqrt{4-3x})dx$$


SOLUTION

We have,

$$I=\int_{0}^{1}{\left( \sqrt{x} \right)dx}+\int_{0}^{\frac{4}{3}}{\left( \sqrt{4-3x} \right)}dx$$


Now,

$$ \int_{0}^{1}{{{\left( x \right)}^{\frac{1}{2}}}dx}+\int_{0}^{\frac{4}{3}}{{{\left( 4-3x \right)}^{\frac{1}{2}}}}dx $$

$$ \Rightarrow {{\left[ \dfrac{{{x}^{\frac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]}_{0}}^{1}+{{\left[ \dfrac{{{\left( 4-3x \right)}^{\frac{1}{2}+1}}}{\left( \dfrac{1}{2}+1 \right)\times -3} \right]}_{0}}^{\frac{4}{3}} $$

$$ \Rightarrow {{\left[ \dfrac{{{x}^{\frac{3}{2}}}}{\dfrac{3}{2}} \right]}_{0}}^{1}-{{\left[ \dfrac{{{\left( 4-3x \right)}^{\frac{3}{2}}}}{3\times \dfrac{3}{2}} \right]}_{0}}^{4/3} $$

$$ \Rightarrow {{\left[ \dfrac{2}{3}{{x}^{\frac{3}{2}}} \right]}_{0}}^{1}-{{\left[ \dfrac{2{{\left( 4-3x \right)}^{\frac{3}{2}}}}{9} \right]}_{0}}^{4/3} $$

$$ \Rightarrow \dfrac{2}{3}\left( {{1}^{\frac{3}{2}}}-{{0}^{\frac{3}{2}}} \right)-\dfrac{2}{9}\left[ {{\left( 4-3\times \dfrac{4}{3} \right)}^{\frac{3}{2}}}-{{\left( 4-3\times 0 \right)}^{\frac{3}{2}}} \right] $$

$$ \Rightarrow \dfrac{2}{3}\times 1-\dfrac{2}{9}\left[ 0-{{4}^{\frac{3}{2}}} \right] $$

$$ \Rightarrow \dfrac{2}{3}-\dfrac{2}{9}\left( -8 \right) $$

$$ \Rightarrow \dfrac{2}{3}+\dfrac{16}{9} $$

$$ \Rightarrow \dfrac{22}{9} $$


Hence, this is the answer.

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