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"#