Advertisements
Advertisements
Question
Advertisements
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:
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
