Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} d x . Then, \]
\[I = \int_0^2 \frac{1}{\sqrt{- x^2 + 2x - 1 + 1 + 3}} d x\]
\[ \Rightarrow I = \int_0^2 \frac{1}{\sqrt{- \left( x - 1 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 1 \right)}{2} \right]_0^2 \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} \left( - \frac{1}{2} \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} \frac{1}{2}\]
\[ \Rightarrow I = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
Evaluate each of the following integral:
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
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
Which notation represents a definite integral?
