मराठी

1 ∫ − 1 5 X 4 √ X 5 + 1 D X - Mathematics

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
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.2 | Q 39 | पृष्ठ ३९

संबंधित प्रश्‍न

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

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

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]

 


Evaluate the following integral:

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

Evaluate the following integral:

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

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

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

 


\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

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

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

\[\int\limits_0^2 e^x dx\]

\[\int\limits_a^b e^x dx\]

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

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

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

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

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

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

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

\[\int\limits_0^{2a} f\left( x \right) dx\]  is equal to


If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


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


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


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


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


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Choose the correct alternative:

Γ(n) is


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×