Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_0^2 x\sqrt{2 - x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{2 - 2 + x} d x\]
\[ = \int_0^2 \left( 2 - x \right)\sqrt{x} d x\]
\[ = \int_0^2 \left( 2\sqrt{x} - x\sqrt{x} \right) dx\]
\[ = \int_0^2 \left( 2 x^\frac{1}{2} - x^\frac{3}{2} \right) dx\]
\[ = \left[ 2\frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^\frac{5}{2}}{\frac{5}{2}} \right]_0^2 \]
\[ = \left[ \frac{4}{3} x^\frac{3}{2} - \frac{2}{5} x^\frac{5}{2} \right]_0^2 \]
\[ = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \]
`=(5xx8sqrt2)/(3xx5)-(3xx8sqrt2)/(5xx3)`
`=(16sqrt2)/15`
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
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:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
