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,

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

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

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

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

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

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

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

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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle \int \frac{1}{x\sqrt{1-x^{3}}}dx=a\log \left | \frac{\sqrt{1-x^{3}}-1}{\sqrt{1-x^{3}}+1} \right |+b$, then a is equal to
• A. $\dfrac 23$
• B. $-\dfrac 13$
• C. $-\dfrac 23$
• D. $\dfrac 13$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate $\displaystyle\int_{0}^{1}\dfrac{2x}{1+x^{4}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int\frac{(sin^{-1}x)^{3}}{\sqrt{1-x^{2}}}dx=$
• A. $\displaystyle \frac{(sin^{-1}x)^{3}}{3}+c$
• B. $(sin^{-1}x)^{4}+c$
• C. $(sin^{-1}x)^{3}+c$
• D. $\displaystyle \frac{1}{4}(sin^{-1}x)^{4}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Integrate
$\int {{x \over {{x^2} + x + 1}}dx}$

$\displaystyle\int \left(e^x\right)^2 e^x dx$ is equal to