1. ## Area of shaded region

I posted a question similar to this a minute ago, but I wanted to double check and make sure I'm doing it right. The question is find the shaded region of $7x(x^2-64)$ There is a picture, but I can't display it here. The bounds are 0 to 8 and the second equation is y = 0. So, you take the top curve (0) - the bottom curve $7x(x^2-64)$. Leaving you with $-7x^3-448x$. From there you integrate which gives $-7/4x^4 - 224x^2$. This is where I get confused with definite integration. Do I put in 8 for the first x and then 0 for the second x? E.g. $-7/4(8)^4 - 224(0)^2$ Or do I do two whole separate equations E.g. $(-7/4(8)^4 - 224(8)^2) - (-7/4(0)^4 - 224(0)^2)$ . Can someone explain? Thanks

2. Originally Posted by zsig013
I posted a question similar to this a minute ago, but I wanted to double check and make sure I'm doing it right. The question is find the shaded region of $7x(x^2-64)$ There is a picture, but I can't display it here. The bounds are 0 to 8 and the second equation is y = 0. So, you take the top curve (0) - the bottom curve $7x(x^2-64)$. Leaving you with $-7x^3-448x$. From there you integrate which gives $-7/4x^4 - 224x^2$. This is where I get confused with definite integration. Do I put in 8 for the first x and then 0 for the second x? E.g. $-7/4(8)^4 - 224(0)^2$ Or do I do two whole separate equations E.g. $(-7/4(8)^4 - 224(8)^2) - (-7/4(0)^4 - 224(0)^2)$ . Can someone explain? Thanks
Remember to evaluate the definite integral, we need to use the Fundamental Theorem of Calculus:
$\int_{a}^{b}f(x)\,dx=F(b)-F(a)$
where $F(x)$ is the antiderivative of $f(x)$.

In your case, we have the antiderivative $-\frac{7}{4}x^4-224x^2$. Then we need to find $F(8)-F(0)=(-\frac{7}{4}(8)^4-224(8)^2)-(-\frac{7}{4}(0)^4-224(0)^2)$.

I hope this clarified things!!!