Advertisements
Advertisements
Question
Evaluate the following integral:
`int_0^1 sqrt(x(x - 1)) "d"x`
Advertisements
Solution
`int_0^1 sqrt(x(x - 1)) "d"x = int_0^1 sqrt(x^2 - x) "d"x`
= `int_0^1 sqrt((x - 1/2)^2 - (1/2)^2)`
This is of the form `int_0^2 sqrt(x^2 - "a"^2)`
= `[((x - 1/2))/2 sqrt((x - 1/2)^2 - (1/2)^2) = 1/((4)(2)) log|(x - 1/2) + sqrt(x(x - 1))|]_0^1`
= `((1 - 1/2))/2 sqrt((1 - 1/2)^2 - (1/2)^2) - 1/8 log|(1 - 1/2) + sqrt(1(1 - 1))|`
= `(-((-1)/2))/2 sqrt(((-1)/2)^2 - (1/2)^2) + 1/8 log|(-1)/2 + sqrt(0(0 - 1))|`
= `1/4 (0) - 1/8 log 1/2 + 1/4 (0) - 1/8 log 1/2`
= 0
APPEARS IN
RELATED QUESTIONS
Integrate the following with respect to x.
`(x^4 - x^2 + 2)/(x - 1)`
Integrate the following with respect to x.
`("e"^x + 1)^2 "e"^x`
Integrate the following with respect to x.
`[1 - 1/2]"e"^((x + 1/x))`
Integrate the following with respect to x.
x log x
Integrate the following with respect to x.
`1/(2x^2 - 9)`
Choose the correct alternative:
`int (sin5x - sinx)/(cos3x) "d"x` is
Choose the correct alternative:
`int sqrt("e"^x) "d"x` is
Choose the correct alternative:
`int "e"^x/("e"^x + 1) "d"x` is
Choose the correct alternative:
`int (2x^3)/(4 + x^4) "d"x` is
Evaluate the following integral:
`int_(-1)^1 x^2 "e"^(-2x) "d"x`
