Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^2 \sqrt{4 - x^2} d x\]
\[ = \int_0^2 \sqrt{2^2 - x^2} d x\]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{1}{2} \times 2^2 \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = \left[ \frac{x}{2}\sqrt{4 - x^2} \right]_0^2 + 2 \left[ \sin^{- 1} \frac{x}{2} \right]_0^2 \]
\[ = 0 + 2\left( \frac{\pi}{2} - 0 \right)\]
\[ = \pi\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Prove that:
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) 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 : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
