Integrals Clarification Needed 
Missed a couple days of class, can't seem to figure out how to get the numbers marked with the blue arrows. Other than that I understand the concept.
 A: Alright, I've done this problem on ms paint. This is how the answer is $-136$
We have the definite integral of 
$\int^4_2 (10+4x-3x^3)\,dx$
Before we can evaluate the definite integral, we need to take the antiderivative of $10+4x-3x^3$. To take the antiderivative we need to add one to the exponent and divide by the new exponent number. Since we only have $10$, the antiderivative is $10x$. 
How did we get $10x?$
Well, when we had $10$, there was no exponent, so when we add the exponent $x$ that's $x^1$ but we don't write it that often, and divide by the new exponent number, so we have $\frac{10x^1}{1} \rightarrow 10x$
For $4x$, we add one to the exponent so that becomes $4x^{1+1}$. Since we have $4x^2$, we need to divide by the new exponent number which is 2. $\frac{4x^2}{2}$ and that becomes $2x^2$.
For $3x^3$ the new exponent number is $4$ and we divide by $4$. There is nothing to reduce, so we're done. 
$3x^{3+1} \rightarrow 3x^4 \rightarrow \frac{3x^4}{4}$
So now we have $10x+2x^2 -\frac{3x^4}{4}$
You can check to see if it's correct by taking the derivative, and it will be $10 +4x-3x^3$
$\int^4_2 (10x+2x^2-\frac{3x^4}{4})\,dx$
Now we evaluate using $F(b) -F(a)$ where $b=4 $ and $ a=2$.
For $F(4)-F(2)$, we have
$10(4)+2(4)^2-\frac{3(4)^4}{4}-[10(2)+2(2)^2-\frac{3(2)^4}{4}]$
$40+2(16)-\frac{256(3)}{4}-[20+8-\frac{48}{4}]$
$40+32-\frac{768}{4}-[20+8-12]$
$72-192-[28-12]$
$72-192-28+12$
$-148+12$
$-136$
