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What is `int_0^(pi/2)` (3cosx-7sinx)dx?

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ruals | (Level 1) Salutatorian

Posted September 12, 2013 at 4:40 PM via web

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What is `int_0^(pi/2)` (3cosx-7sinx)dx?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 12, 2013 at 5:25 PM (Answer #1)

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You need to use the property of linearity of integrals, such that:

`int_0^(pi/2) (3cos x - 7 sin x)dx = int_0^(pi/2) (3cos x) dx - int_0^(pi/2) (7 sin x)dx`

Taking out the constants yields:

`int_0^(pi/2) (3cos x - 7 sin x)dx = 3 int_0^(pi/2) (cos x) dx - 7int_0^(pi/2) (sin x)dx`

`int_0^(pi/2) (3cos x - 7 sin x)dx = 3 sin x|_0^(pi/2) - 7*(-cos x)|_0^(pi/2) `

By fundamental theorem of calculus, yields:

`int_0^(pi/2) (3cos x - 7 sin x)dx = 3(sin (pi/2) - sin 0) + 7(cos (pi/2) - cos 0)`

`int_0^(pi/2) (3cos x - 7 sin x)dx = 3(1 - 0) + 7(0 - 1)`

`int_0^(pi/2) (3cos x - 7 sin x)dx = 3 - 7`

`int_0^(pi/2) (3cos x - 7 sin x)dx = -4`

Hence, evaluating the given definite integral, using the property of linearity of integrals, yields `int_0^(pi/2) (3cos x - 7 sin x)dx = -4.`

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