Answer
$\frac{17}{2}$
Work Step by Step
Evaluate the Integral: $\int^{2}_{1}\frac{v^5+3v^6}{v^4}dv $
Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$
Find $F(x)$:
$F(x) = \int\frac{v^5 +3v^6}{v^4}dv$
$F(x) = \int \frac{v^5}{v^4}dv +\int\frac{3v^6}{v^4}dv$
$F(x) = \int{vdv} + \int3v^2dv$
$F(x) = \frac{v^2}{2} + \frac{3v^3}{3}$
$F(x) = \frac{v^2}2 + v^3 $
Now Evaluate: $F(b) - F(a)$
$F(2) - F(1)$
$(\frac{(2)^2}{2}+(2)^3) - (\frac{(1)^2}{2} + (1)^3)$
$(2 + 8) - (\frac{1}{2}+1)$
$10 - \frac{3}{2}$
$\frac{17}{2}$
