English

Evaluate: ∫-12|x3-3x2+2x|dx - Mathematics

Advertisements
Advertisements

Question

Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`

Sum
Advertisements

Solution

The given definite integral = `int_(-1)^2|x(x - 1)(x - 2)|dx`

= `int_(-1)^0 |x(x - 1)(x - 2)|dx + int_0^1 |x(x - 1)(x - 2)|dx + int_1^2 |x(x - 1)(x - 2)|dx`

= `- int_(-1)^0 (x^3 - 3x^2 + 2x)dx + int_0^1 (x^3 - 3x^2 + 2x)dx - int_1^2 (x^3 - 3x^2 + 2x)dx`

= `- [x^4/4 - x^3 + x^2]_(-1)^0 + [x^4/4 - x^3 + x^2]_0^1 - [x^4/4 - x^3 + x^2]_1^2`

= `9/4 + 1/4 + 1/4 = 11/4`

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
2021-2022 (March) Term 2 Sample

RELATED QUESTIONS

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos\ x\ dx\]

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

\[\int\limits_0^{\pi/2} \sin^3 x\ dx\]

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\]

 


\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_0^2 x\sqrt{2 - x} dx\]

\[\int\limits_0^2 \left( x^2 + 1 \right) dx\]

\[\int\limits_0^{\pi/2} \cos x\ dx\]

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_0^1 \frac{1}{x^2 + 1} dx\]

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]


\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

`Γ(3/2)`


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×