Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} d x . Then, \]
\[I = \int_0^2 \frac{1}{\sqrt{- x^2 + 2x - 1 + 1 + 3}} d x\]
\[ \Rightarrow I = \int_0^2 \frac{1}{\sqrt{- \left( x - 1 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 1 \right)}{2} \right]_0^2 \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} \left( - \frac{1}{2} \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} \frac{1}{2}\]
\[ \Rightarrow I = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate each of the following integral:
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
Γ(n) is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
