Evaluate the integral int_1^2 dt/(8-3t)?
1 Answer
Mar 15, 2017
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"