Advertisements
Advertisements
Question
Evaluate `∫_0^(3/2)|x cosπx|dx`
Advertisements
Solution
`int_0^(3/2)|xcospix|dx`
`0<x<1/2`
`0<pix<pi/2rArrcospix>0rArr(xcospix)>0`
`|xcospix|=xcospix`
`1/2<x<3/2`
`pi/2<pix<(3pi)/2rArrcospix<0rArr(xcospix)<0`
`|xcospix|=-xcospix`
`I=int_0^(3/2)|xcospix|dx=int_0^(3/2)xcospix+int_(1/2)^(3/2)-(xcospix)`
`I=int_0^(1/2)xcospix-int_(1/2)^(3/2)xcospix`
`intx(cospix)=x(sinpix)/pi-int(sinpix)/pi`
`=x/pi(sinpix)+(cospix)/pi^2`
`I=[(x/pisinpix)+(cospix)/pi^2]_0^(1/2)-[(x/pisinpix)+(cospix)/pi^2]_(1/2)^(3/2)`
`=[1/pi((1/2)-0)+1/pi^2(0-1)]-[1/pi(3/2(-1)-1/2(1))+1/pi^2(0-0)]`
`=(1/(2pi)-1/pi^2)-((-2)/pi)`
`=(5/(2pi)-1/pi^2)`
`=((5pi-2)/(2pi^2))`
RELATED QUESTIONS
Evaluate `int_(-1)^2|x^3-x|dx`
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
Evaluate :
`int_e^(e^2) dx/(xlogx)`
Evaluate: `intsinsqrtx/sqrtxdx`
Evaluate the integral by using substitution.
`int_0^2 dx/(x + 4 - x^2)`
Evaluate `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`
Evaluate of the following integral:
Evaluate:
Evaluate:
Evaluate the following integral:
\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Find : \[\int e^{2x} \sin \left( 3x + 1 \right) dx\] .
Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .
`int_0^3 1/sqrt(3x - x^2)"d"x` = ______.
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
Evaluate the following:
`int ("e"^(6logx) - "e"^(5logx))/("e"^(4logx) - "e"^(3logx)) "d"x`
Evaluate the following:
`int "dt"/sqrt(3"t" - 2"t"^2)`
`int_0^1 x^2e^x dx` = ______.
Under Method: Changing the Limits, what does the lower limit \[a\] become?
For the integral \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which substitution is suitable?
What is the value of \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\]?
