Evaluate the integral int_1^2 dt/(8-3t)?

1 Answer
Mar 15, 2017

1/3ln(5/2)~~0.305" to 3 decimal places"

Explanation:

Using the standard result for integrals of the form.

color(red)(bar(ul(|color(white)(2/2)color(black)(int((f'(x))/(f(x)))dx=ln|f(x)|+c)color(white)(2/2)|)))

f(t)=8-3trArrf'(t)=-3

rArrint_1^2(dt)/(8-3t)

=-1/3int_1^2(-3)/(8-3t)dt

=-1/3[ln(8-3t)]_1^2

-1/3(ln2-ln5)

1/3(ln5-ln2)

=1/3ln(5/2)~~0.305" to 3 decimal places"