Advertisements
Advertisements
Question
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
Advertisements
Solution
\[Let, I = \int_0^4 x\sqrt{4 - x} d x\]
\[ = \int_0^4 \left( 4 - x \right)\sqrt{4 - 4 + x} d x\]
\[ = \int_0^4 \left( 4 - x \right)\sqrt{x} d x\]
\[ = \int_0^4 4\sqrt{x} - x^\frac{3}{2} dx\]
\[ = \left[ 8\frac{x^\frac{3}{2}}{3} \right]_0^4 - \left[ \frac{2 x^\frac{5}{2}}{5} \right]_0^4 \]
\[ = \frac{64}{3} - \frac{64}{5}\]
\[ = \frac{128}{15}\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Which integral has lower and upper limits?
