Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle \lim_{n\rightarrow \infty }\frac{1^{p}+2^{p}+...+n^{p}}{n^{p+1}}$ is
• A. $\displaystyle \frac{1}{1-p}$
• B. $\displaystyle \frac{1}{p}-\frac{1}{p-1}$
• C. $\displaystyle \frac{1}{p+2}$
• D. $\displaystyle \frac{1}{p+1}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle \int_{0}^{2}2x dx=$

1 Verified Answer | Published on 17th 09, 2020

Q3 One Word Hard
If $\displaystyle \int_{0}^{1}\frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}=\frac{\pi A}{\sqrt{6}(\sqrt{3}+1) (157)}$, then $A$ is equal to

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle \int _0^{\pi/2} \sin x \cos x dx$

• A. $\dfrac 34$
• B. $2$
• C. $None\ of\ these$
• D. $\dfrac 12$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int_1^{32}\dfrac{dx}{x^{1/5}\sqrt{1+x^{4/5}}}$
• A. $\dfrac{2}{5}(\sqrt{17}-\sqrt{2})$
• B. $\dfrac{5}{2}(\sqrt{17}-\sqrt{2})$
• C. $\dfrac{5}{2}(\sqrt{17}+\sqrt{2})$
• D. $\dfrac{2}{5}(\sqrt{17}+\sqrt{2})$