So first off, I know that I have:
$y=3x^2$ and $y=24-6x$
Do I first set these equations equal to each other? If so, then I get:
$(3x^2)+6x-24=0$.
Therefore, I would get $x=-4$ or $x=-2$
Is this correct so far?
If so, then I know I would put these numbers over and below the integral symbol.
If all of this is correct so far, then my biggest question is this; How do I know which numbers to put above and below the integral symbol, and how do I know what equation to put into the integral function part, in order to take the anti derivative and find the area?
Disclaimer: I am new to this online community, and so if you see something in my question/post that should not belong, or that you do not like, please tell me before down-voting my post. I am not trying to get someone to strictly do homework for me. I am trying to learn so that I can replicate it on similar problems. I want to be respectful. My class is online, so asking specific questions in class is a challenge, as there are no interactive lectures.
So if you want area between the curve and the x-axis (i.e., the line $y = 0$), you need to integrate the following two integrals, and in each, we take the top curve and subtract the bottom curve. For the first integral, $-\sin x > 0$ so the integrand is $-\sin x - 0$, and in the second integral, $0 > -\sin x$, so we need for the integrand to be $0 - (-\sin x)$:
Best Answer
You didn't solve the quadratic equation correctly: $3x^2=24-6x$ gives you $x=-4$ or $x=2$. Then you can check that the two curves meet at $(2,12)$.
In general, the definite integral $\int_a^b f(x)\ dx$ gives you the area of the region under the curve provided by $y=f(x)$. Here, you have $a=0$, $b=4$ and $$ f(x)= \begin{cases} 3x^2,& 0\leqslant x\leqslant 2,\\ 24-6x, & 2\leqslant x\leqslant 4. \end{cases} $$ Remember that integration is "additive": $$ \int_{a}^bf(x)\ dx = \int_a^cf(x)\ dx+\int_c^bf(x)\ dx. $$ Here you have $c=2$.