Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_0^2 \sqrt{4 - x^2} d x\]
\[ = \int_0^2 \sqrt{2^2 - x^2} d x\]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{1}{2} \times 2^2 \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} \right]_0^2 + 2 \left[ \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = 0 + 2\left( \frac{\pi}{2} - 0 \right)\]
\[ = \pi\]
APPEARS IN
RELATED QUESTIONS
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^4 x dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
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:
Γ(4)
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
