1. Area between 2 graphs

Find the area enclosed between and from to .

$\int_{-2}^{6} [(0.4^2+3) - (x)] dx$

$\frac {0.4x^3}{3} + 3x - \frac {x^2}{2}$

If I'm not mistaken, I now have to use the Fb - Fa formula. However, my answer turns out wrong so I'm assuming I made a mistake in the calculations shown above. Can someone please tell me where it is?

2. Originally Posted by Archduke01
Find the area enclosed between and from to .

$\int_{-2}^{6} [(0.4^2+3) - (x)] dx$

$\frac {0.4x^3}{3} + 3x - \frac {x^2}{2}$

If I'm not mistaken, I now have to use the Fb - Fa formula. However, my answer turns out wrong so I'm assuming I made a mistake in the calculations shown above. Can someone please tell me where it is?
your antiderivative is correct ... error in evaluation from x = -2 to x = 6, maybe?

3. $\frac {0.4x^3}{3} + 3x - \frac {x^2}{2}$

$b = 6$
$\frac {0.4*6^3}{3} + 3*6 - \frac {6^2}{2} = 28.8$

$a = -2$
$
\frac {0.4*(-2)^3}{3} + 3*(-2) - \frac {(-2)^2}{2} = -5.07$

$F(b) - F(a) = 28.8 - (-5.07) = 33.86$

I don't understand ... I've triple-checked, and I get the same answer every time. It's wrong.

4. $\frac{0.4(-2)^3}{3} + 3(-2) - \frac{(-2)^2}{2}$

$\frac{0.4(-8)}{3} - 6 - \frac{4}{2}$

$\frac{-3.2}{3} - 6 - 2 = -9.0666...$