Monday, August 17, 2015

Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 4

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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