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
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
If f is an integrable function, show that
Evaluate each of the following integral:
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
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:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
