Mathematics

# Solve: $\displaystyle \int _{ -1 }^{ 1 }{ \dfrac { 2\sin { x-3{ x }^{ 2 } } }{ 4-\left| x \right| } dx= }$

$6\int _{ 0 }^{ 1 }{ \dfrac { { x }^{ 2 } }{ 4-\left| x \right| } dx }$

##### SOLUTION

consider, $\displaystyle I= \int_{ - 1}^1 {\frac{{2sinx - 3{x^2}}}{{4 - |x|}}dx}$

$\displaystyle I=\int_{ - 1}^0 {\dfrac{{2sinx - 3{x^2}}}{{4 - |x|}}dx} + \int_0^1 {\dfrac{{2sinx - 3{x^2}}}{{4 - |x|}}dx}$

$\displaystyle I=\int_{ - 1}^0 {\dfrac{{2sinx - 3{x^2}}}{{4 + x}}dx} + \int_0^1 {\dfrac{{2sinx - 3{x^2}}}{{4 - x}}dx}$

putting $x=-y,\;dx=-dy$

So,$x \rightarrow -1 \ \Rightarrow y \rightarrow 1$ and $x \rightarrow 0 \ \Rightarrow y \rightarrow 0$

$\Rightarrow \displaystyle I= \int_1^0 {\dfrac{{2\sin y + 3{y^2}}}{{4 - y}}dy} + \int_0^1 {\dfrac{{2{\mathop{\rm sinx}\nolimits} - 3{x^2}}}{{4 - x}}dx}$

$\displaystyle I= - \int_0^1 {\dfrac{{2\sin y + 3{y^2}}}{{4 - y}}dy} + \int_0^1 {\frac{{2{\mathop{\rm sinx}\nolimits} - 3{x^2}}}{{4 - x}}dx}$

$\displaystyle I= - \int_0^1 {\dfrac{{2{\mathop{\rm sinx}\nolimits} + 3{x^2}}}{{4 - x}}dx} + \int_0^1 {\dfrac{{2{\mathop{\rm sinx}\nolimits} - 3{x^2}}}{{4 - x}}dx}$                      $\because \left[ {\int_0^a {f\left( t \right)dt} = \int_0^a {f\left( x \right)dx} } \right]$

$\displaystyle I= \int_0^1 {\dfrac{{ - 2{\mathop{\rm sinx}\nolimits} - 3{x^2} + 2{\mathop{\rm sinx}\nolimits} - 3{x^2}}}{{4 - x}}dx}$

$\displaystyle I= \int_0^1 {\dfrac{{ - 6{x^2}}}{{4 - x}}dx}$

$\displaystyle I= - 6\int_0^1 {\dfrac{{{x^2}}}{{4 - |x|}}dx}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Medium
$\int \sin x . \cos x . \cos 2x . \cos 4x . \cos 8x . \cos 16x\ dx$ equals.
• A. $\dfrac {\sin 16x}{1024} + c$
• B. $\dfrac {\cos 32x}{1096} + c$
• C. $-\dfrac {\cos 32x}{1096} + c$
• D. $\dfrac {-\cos 32x}{1024} + c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Multiple Correct Hard
$\displaystyle\int{\frac{\sqrt{4+x^2}}{x^6}dx}=\frac{A{(4+x^2)}^{\frac{3}{2}}(Bx^2-6)}{x^5}+C$
• A. $\displaystyle A=-\frac{1}{120}$
• B. $B=-1$
• C. $\displaystyle A=\frac{1}{120}$
• D. $B=1$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate:

$\displaystyle \int\dfrac{\sin^6x+\cos^6x}{\sin^2x \cos^2x}dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate:$\displaystyle\int{\dfrac{x\,dx}{a+bx}}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
If $A=\displaystyle \int_{0}^{1}{\dfrac{{e}^{t}}{1+t}}dt$ then $\displaystyle \int_{0}^{1}{{e}^{t}ln(1+t)}dt=$
• A. $e\ ln{2}+A$
• B. $Ae\ ln{2}$
• C. $A\ ln{2}$
• D. $e\ ln{2}-A$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020