$\displaystyle [0(lower),2(upper)]\int 5x~dx$
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Originally Posted by katchat64 $\displaystyle [0(lower),2(upper)]\int 5x~dx$ Recall that $\displaystyle \int{ax^n\,dx} = \frac{a}{n + 1}x^{n + 1} +C$ And if $\displaystyle F(x) = \int{f(x)\,dx}$ then $\displaystyle \int_a^b{f(x)\,dx} = F(b) - F(a)$. Just substitute the correct values.
I thought that you were suppose to take the antiderivaitve of 5x and then substitute in 0 & 2. Is that what you meant?
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.
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: $\displaystyle \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 $
Well that kinda got me confused, but it doesn't hurt to know the terminology. Thanks.
I have one more question for you. How did you get the numbers on the integral sign, the ones that tell you the range.
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} where a is your lower limit and b is your upper.
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