Answer to Question #99230 in Calculus for Leo

Question

Questions: A) find the indefinite integral of the function y = 3t^2 + 2e^3t + 1/t + 2cos3t?
B) calculate the definite integral: (this is an integral operator, with a 1 at the bottom and 2 at the top of the integral operator) 1int2 3t^2 + 2e^3t + 1/t + 2cos3t dt ?

Answer 1

A) $\int 3t^2 + 2e^{3t} + \cfrac{1}{t} + 2\cos3t\,dt = t^3 + \cfrac{2}{3}e^{3t} + \ln|t| + \cfrac{2}{3}\sin3t + C$

B)

$\int_1^2 3t^2 + 2e^{3t} + \cfrac{1}{t} + 2\cos3t\,dt = \Big[t^3 + \cfrac{2}{3}e^{3t} + \ln\,t + \cfrac{2}{3}\sin3t\Big]_1^2 =$

$2^3 - 1^3 + \cfrac{2}{3}(e^{3*2}-e^{3*1}) + ln2-ln1 +\cfrac{2}{3}(\sin(3*2) - \sin(3*1)) =$

$7 + \cfrac{2}{3}(e^6 - e^3+\sin6 - \sin3) +\ln2 \approx 262.97$