मराठी

1 ∫ 0 1 1 + X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let I = \int_0^1 \frac{1}{1 + x^2} d x . Then, \]
\[I = \left[ \tan^{- 1} x \right]_0^1 \]
\[ \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{4} - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.1 | Q 4 | पृष्ठ १६

संबंधित प्रश्‍न

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

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

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

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

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

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

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

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

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

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

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

 


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

 


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

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

 


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

 


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

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

 


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

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

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

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

\[\int\limits_0^{\pi/4} \tan^2 x\ dx .\]

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

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

\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]


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


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


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


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Find : `∫_a^b logx/x` dx


Evaluate the following:

`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×