Mathematics

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


ANSWER

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


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 $$x \rightarrow 0 \ \Rightarrow y \rightarrow 0$$


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


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


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


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


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


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


View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84
Enroll Now For FREE

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

View Answer
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

View Answer
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

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

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
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

View Answer