Mathematics

# Solve : $\displaystyle\int_{-2}^{2}|2x+3|\ dx$

##### SOLUTION
Let $I=\displaystyle \int_{-2}^{2}|2x+2|dx$

we know
$|2x+3|=\begin{cases} -\left( 3x+3 \right) & x<\dfrac { -3 }{ 2 } \\ 2x+3 & x\ge \dfrac { 3 }{ 2 } \end{cases}$

then

$I=I=\displaystyle \int_{-2}^{-3/2} -2x-3\ dx +I=\displaystyle \int_{-3/2}^{2} 2x+3\ dx$

$=-x^2 -3x|_{-2}^{-3/2} +x^2 +3x|_{-3/2}^2$

$=\left (-\dfrac {9}{4}+\dfrac {9}{2}\right)-(5+6)+(4+6)-\left (\dfrac {9}{4}-\dfrac {9}{2}\right)$

$=\dfrac {9}{2}-2+10+\dfrac {9}{4}=8+\dfrac {9}{2}$

$=\dfrac {25}{2}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Subjective Hard
$\int {\dfrac {(x+1)}{x(1+xe^{x})^{2}}}dx$ is equal to

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int_{0}^{1}\frac{3^{x+1}-4^{x-1}}{12^{x}}dx=$

• A. $\displaystyle \frac{9}{4}log_{4}\, e$
• B. $\displaystyle log_{4}3$
• C. None of these
• D. $\displaystyle \frac{9}{4}log_{4}\, e-\frac{1}{6}log_{3}e$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate:
$\int { \sin ^{ 2 }{ x } } \cos ^{ 2 }{ x } dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Show that $\displaystyle \int \frac{dx}{\sqrt{\left (1+e^{x}+e^{2x} \right )}}=\log \left \{ \left ( z+\frac{1}{2} \right )+\sqrt{\left ( z^{2}+z+1 \right )} \right \}$. where $z=\displaystyle \frac{1}{e^{x}}$.

Find $\displaystyle \int \dfrac {1}{\sin x\cos^{3}x}dx.$