# `int y e^(0.2y) dy` Evaluate the integral

You need to use the substitution` 0.2y = u` , such that:

`0.2y=  u => 0.2dy = du => dy= (du)/(0.2)`

Replacing the variable, yields:

`int y*e^(0.2y) dy = (10/2)int u/(0.2)*e^u du`

You need to use the integration by parts such that:

`intfdg =fg - int gdf `

`f =u => df = du`

`dg = e^u=> g = e^u`

`25int u*e^u du = 25(u*e^u - int e^u du)`

`25int u*e^u du = 25u*e^u - 25e^u + c`

Replacing back the variable, yields:

`int y*e^(0.2) dy = 25((0.2y)*e^(0.2y) - e^(0.2y)) + c`

Hence, evaluating the integral, using substitution, then integration by parts, yields `int y*e^(0.2) dy = (25(e^(0.2y)))(0.2y - 1) + c`

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You need to use the substitution `0.2y = u` , such that:

`0.2y=  u => 0.2dy = du => dy= (du)/(0.2)`

Replacing the variable, yields:

`int y*e^(0.2y) dy = (10/2)int u/(0.2)*e^u du`

You need to use the integration by parts such that:

`int fdg = fg - int gdf`

`f = u => df = du`

`dg = e^u=> g = e^u`

2`5int u*e^u du = 25(u*e^u - int e^u du)`

`25int u*e^u du = 25u*e^u - 25e^u + c`

Replacing back the variable, yields:

`int y*e^(0.2) dy = 25((0.2y)*e^(0.2y) - e^(0.2y)) + c`

Hence, evaluating the integral, using substitution, then integration by parts, yields `int y*e^(0.2) dy = ((e^(0.2y))/9)(0.2y - 1) + c`

Approved by eNotes Editorial Team

Posted on

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