Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_0^1 \frac{1}{1 + x^2} d x . Then, \]
\[I = \left[ \tan^{- 1} x \right]_0^1 \]
\[ \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{4} - 0\]
\[ \Rightarrow I = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
Evaluate :
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
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
