Answer
$\frac{26}{3}$
Work Step by Step
Evaluate the integral: $\int^3_1 (x^2 +2x -4) dx$
Recall the 2nd part of the Fundamental Theorem of Calculus:
$\int_a^bf(x)dx = F(b) - F(a)$
Find $F(x)$:
$F(x) = \int (x^2 +2x-4)dx $
$F(x) = \frac{x^3}{3} + \frac{2x^2}{2} -\frac{4x^1}{1} $
$F(x) =\frac{x^3}{3}+x^2 -4x $
Now evaluate $F(b) - F(a)$:
$F(3) - F(1)$
$(\frac{(3)^3}{3}+(3)^2 -4(3) )- (\frac{(1)^3}{3}+(1)^2 -4(1)) $
$(9+9 -12) - (\frac{1}{3} +1 - 4)$
$6 + 3 - \frac{1}{3}$
$\frac{27}{3} -\frac{1}{3} = \frac{26}{3}$
