हिंदी

Evaluate the Following Integral: 2 ∫ − 2 | 2 X + 3 | D X

Advertisements
Advertisements

प्रश्न

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]
योग
Advertisements

उत्तर

\[\int_{- 2}^2 \left| 2x + 3 \right| d x\]
\[We\ know\ that\ \left| 2x + 3 \right| = \begin{cases} - \left( 2x + 3 \right) &, &- 2 \leq x \leq - \frac{3}{2}\\\left( 2x + 3 \right)&, &- \frac{3}{2} < x \leq 2\end{cases}\]
\[ \therefore I = \int_{- 2}^\frac{- 3}{2} - \left( 2x + 3 \right) d x + \int_{- \frac{3}{2}}^2 \left( 2x + 3 \right) d x\]
\[ \Rightarrow I = - \left[ x^2 + 3x \right]_{- 2}^\frac{- 3}{2} + \left[ x^2 + 3x \right]_{- \frac{3}{2}}^2 \]
\[ \Rightarrow I = - \frac{9}{4} + \frac{9}{2} + 4 - 6 + 4 + 6 - \frac{9}{4} + \frac{9}{2}\]
\[ \Rightarrow I = \frac{25}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.3 | Q 5 | पृष्ठ ५६

संबंधित प्रश्न

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

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

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

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

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

\[\int_0^1 | x\sin \pi x | dx\]

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

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

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

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

\[\int\limits_{- \pi/2}^{\pi/2} x \cos^2 x\ dx .\]

 


\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

\[\int\limits_2^3 \frac{1}{x}dx\]

Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\] equals

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .

 

\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


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


What is an antiderivative of \[x\]?


Evaluate \[\int_{1}^{3}(2x+1)\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×