मराठी

∫ 2 − 1 ( | X + 1 | + | X | + | X − 1 | ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 

बेरीज

उत्तर

We know that

\[\left| x + 1 \right| = \begin{cases}x + 1, & \text{if }x + 1 \geq 0 \\ - \left( x + 1 \right), & \text{if }x + 1 < 0\end{cases} = \begin{cases}x + 1, & \text{if }x \geq - 1 \\ - \left( x + 1 \right), & \text{if }x < - 1\end{cases}\]

\[\left| x \right| = \begin{cases}x, & \text{if }x \geq 0 \\ - x, & \text{if }x < 0\end{cases}\]

\[\left| x - 1 \right| = \begin{cases}x - 1, & \text{if }x - 1 \geq 0 \\ - \left( x - 1 \right), & \text{if }x - 1 < 0\end{cases} = \begin{cases}x - 1, & \text{if }x \geq 1 \\ - \left( x - 1 \right), & \text{if }x < 1\end{cases}\]

When

\[- 1 \leq x \leq 0,\]
\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + \left( - x \right) + \left[ - \left( x - 1 \right) \right] = 2 - x\]

When

\[0 \leq x \leq 1,\]
\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + x + \left[ - \left( x - 1 \right) \right] = x + 2\]

When

\[1 \leq x \leq 2,\]
\[\left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| = x + 1 + x + x - 1 = 3x\]

\[\therefore \int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]
\[ = \int_{- 1}^0 \left( 2 - x \right)dx + \int_0^1 \left( x + 2 \right)dx + \int_1^2 3xdx\]
\[ = \left.\frac{\left( 2 - x \right)^2}{2 \times \left( - 1 \right)}\right|_{- 1}^0 + \left.\frac{\left( x + 2 \right)^2}{2}\right|_0^1 + \left.3 \times \frac{x^2}{2}\right|_1^2 \]
\[ = - \frac{1}{2}\left( 4 - 9 \right) + \frac{1}{2}\left( 9 - 4 \right) + \frac{3}{2}\left( 4 - 1 \right)\]
\[ = \frac{5}{2} + \frac{5}{2} + \frac{9}{2}\]
\[ = \frac{19}{2}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.3 | Q 20 | पृष्ठ ५६

संबंधित प्रश्‍न

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

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

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

Evaluate the following integral:

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

Evaluate the following integral:

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

\[\int_{- 2}^2 x e^\left| x \right| dx\]

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 


Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

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

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

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

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

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

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

\[\int\limits_0^{15} \left[ x \right] dx .\]

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

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

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


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

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

Γ(4)


Verify the following:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×