Calc 2 Problem: Improper Integrals

Question

Let R be the region bounded by the graph of y=e^(-ax) and y=e^(-bx) for x is greater than or equal to zero where a>b>0. Find the area of R. 

 

Any help with this would be much appreciated! I thought I had the right answer, but I was off. Thank you.

Answer 1

Since a > b > 0, in the interval [0, infinity), the graph of e^−bx will always be above the graph of e^−ax. So the set-up of the area integral is ∫(e^−bx − e^−ax)dx from 0 to ∞. This can be rewritten as the difference of two integrals:

 

(∫e^−bxdx from 0 to ∞) − (∫e^−axdx from 0 to ∞).

 

For the first, we can write lim N→∞ [(−1/b)e^−bx] from 0 to N and for the second lim N→∞ [(−1/a)e^−ax] from 0 to N.

 

Applying the Fundamental Theorem of Calculus:

 

lim N→∞{[(−1/b)(e^−Nb −  e⁰)] − [(−1/a)(e^−Na − e⁰)]} the terms with −Nk go to 0, since e^−Nk = 1/e^Nk → 0 as N→∞, where k is a constant.

 

We have (−1/b)(−1) − (−1/a)(−1) = (1/b) − (1/a) = (a − b)/ab.

Answer 2

Since a>b,given

 

Plot of e^(-bx). shall be above the plot of e^(-ax)

 

We want to find the area in between these two curves.

 

Integral[(e^(-bx)-e^(-ax)]dx         take limits from 0 to infinite.

 

Since this is improper integral

 

Limits M tends to infinite( integral of (e^(-bx)-e^(-ax))         with limits. 0 to M

 

= limits M tends to infinite [ (-1/b)* e^(-bx)+(1/a)*e^(-ax)].       take limits from 0 to M

 

On taking limits

 

(1/b)-(1/a)

 

(a-b)/ab