Answer
2
Work Step by Step
Evaluate the integral: $\int^2_0 (\frac{4}{5}t^3-\frac{3}{4}t^2+\frac{2}{5}t) dx$
Recall the 2nd part of the Fundamental Theorem of Calculus:
$\int_a^bf(x)dx = F(b) - F(a)$
Find $F(x)$:
$F(x) = \int^2_0 (\frac{4}{5}t^3-\frac{3}{4}t^2+\frac{2}{5}t) dx$
$F(x) =\frac{1}{4}(\frac{4}{5}t^4) - \frac{1}{3}(\frac{3}{4}t^3) + \frac{1}{2}(\frac{2}{5}t^2) $
$F(x) = \frac{t^4}{5} - \frac{t^3}{4} +\frac{t^2}{5}$
Now evaluate $F(b) - F(a)$:
$F(2) - F(0)$
$( \frac{2^4}{5} - \frac{2^3}{4} +\frac{2^2}{5}) - ( \frac{0^4}{5} - \frac{0^3}{4} +\frac{0^2}{5})$
$\frac{16}{5} - \frac{8}{4} + \frac{4}{5} - 0$
$ 4 - 2 = 2$