हिंदी

1 ∫ − 1 5 X 4 √ X 5 + 1 D X

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx\]
Advertisements

उत्तर

\[Let\ I = \int_{- 1}^1 5 x^4 \sqrt{x^5 + 1}\ d\ x . Then, \]
\[Let\ x^5 + 1 = t . Then, 5 x^4\ dx = dt\]
\[When\ x = - 1, t = 0\ and\ x = 1, t = 2\]
\[ \therefore I = \int_0^2 \sqrt{t}\ dt\]
\[ \Rightarrow I = \left[ \frac{2}{3} t^\frac{3}{2} \right]_0^2 \]
\[ \Rightarrow I = \frac{2}{3}\sqrt{8}\]
\[ \Rightarrow I = \frac{4\sqrt{2}}{3}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.2 | Q 39 | पृष्ठ ३९

संबंधित प्रश्न

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

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

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

Evaluate the following definite integrals:

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

\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

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

 


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

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

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

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

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

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

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

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

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

\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

\[\int\limits_1^e \log x\ dx =\]

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

Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


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


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


\[\int\limits_2^3 e^{- x} dx\]


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×