1. ## Evaluate the Integral

$[0(lower),2(upper)]\int 5x~dx$

2. Originally Posted by katchat64
$[0(lower),2(upper)]\int 5x~dx$
Recall that $\int{ax^n\,dx} = \frac{a}{n + 1}x^{n + 1} +C$

And if $F(x) = \int{f(x)\,dx}$ then

$\int_a^b{f(x)\,dx} = F(b) - F(a)$.

Just substitute the correct values.

3. I thought that you were suppose to take the antiderivaitve of 5x and then substitute in 0 & 2. Is that what you meant?

4. Originally Posted by katchat64
I thought that you were suppose to take the antiderivaitve of 5x and then substitute in 0 & 2. Is that what you meant?
Yes, that's exactly what I meant. It helps if you learn what the terminology means before you ask how to do something.

5. Originally Posted by katchat64
I thought that you were suppose to take the antiderivaitve of 5x and then substitute in 0 & 2. Is that what you meant?
Yes:

$\int_0^2 5xdx = 5 \int_0^2 x dx = 5 [\frac{x^2}{2}]^2_0 = 5[\frac{2^2}{2} - \frac{0^2}{2}] = 5 \times 2 = 10$

6. Well that kinda got me confused, but it doesn't hurt to know the terminology. Thanks.

7. I have one more question for you. How did you get the numbers on the integral sign, the ones that tell you the range.

8. Originally Posted by katchat64
I have one more question for you. How did you get the numbers on the integral sign, the ones that tell you the range.
The code is

\int_a^b{integrand\,dx}