मराठी

Evaluate the Following Integral: 3 ∫ − 3 | X + 1 | D X

Advertisements
Advertisements

प्रश्न

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]
बेरीज
Advertisements

उत्तर

\[I = \int_{- 3}^3 \left| x + 1 \right| d x\]
\[We\ know\ that, \left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &, &- 3 \leq x \leq - 1 \\x + 1 &, &- 1 < x \leq 3\end{cases}\]
\[ \therefore I = \int_{- 3}^{- 1} - \left( x + 1 \right) d x + \int_{- 1}^3 \left[ x + 1 \right] d x\]
\[ \Rightarrow I = \left[ - \frac{\left( x + 1 \right)^2}{2} \right]_{- 3}^{- 1} + \left[ \frac{\left( x + 1 \right)^2}{2} \right]_{- 1}^3 \]
\[ \Rightarrow I = 0 + 2 + 8 - 0\]
\[ \Rightarrow I = 10\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.3 | Q 3 | पृष्ठ ५६

संबंधित प्रश्‍न

\[\int\limits_0^{\pi/4} \sec x dx\]

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

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

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

\[\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx\]

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

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

\[\int\limits_0^1 x \tan^{- 1} x\ dx\]

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

\[\int_0^\pi \cos x\left| \cos x \right|dx\]

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


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

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

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

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

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

 


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

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

\[\int\limits_0^2 x\left[ x \right] dx .\]

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]


\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]


\[\int\limits_0^{\pi/4} \tan^4 x dx\]


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

Γ(4)


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Evaluate \[\int_{0}^{1}x\,dx\].


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×