How to calculate the integral?

Question

$int_0^(pi/3)$ $sinx * e^(-3x)$ dx

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Answer 1 · 2018-05-05T14:07:20

$I=1/10[1-(3sqrt3+1)/(2e^pi)]$
$I~~-2.214069$

Note:

$color(red)(inte^(ax)sinbxdx=(e^(ax))/(a^2+b^2)(asinbx-bcosbx)+c$

Take, $a=-3and b=1.$

$inte^(-3x)sin(1x)dx=(e^(-3x))/((-3)^2+1^2)(-3sinx- 1cos(1x))+c$

$color(blue)(inte^(-3x)sinxdx=(e^(-3x))/(10)(-3sinx-cosx)+c...to (1)$

Here,

$I=int_0^(pi/3) sinx *e^(-3x)dx$

$I=int_0^(pi/3)(e^(-3x))sinxdx$

Using $(1)$ ,we get

$I=[(e^(-3x))/(10)(-3sinx-cosx)]_0^(pi/3)$

$I=[(e^(-3xxpi/3))/10(-3sin(pi/3)-cos(pi/3)]-[e^0/10(-3(0)-1)]$

$I=[e^-pi/10(-3xxsqrt3/2-1/2)]-[1/10(-1)]$

$I=e^-pi/10(-(3sqrt3+1)/2)+1/10$

$I=1/10[1-(3sqrt3+1)/(2e^pi)]$

$I~~1/10[1-23.14069]$

$I~~1/10[-22.14069]$

$I~~-2.214069$