Evaluating a simple definite integral I'm currently teaching myself calculus and am onto the Mean Value Theorem for Integration.
I am finding the value of $f(c)$ on the function $f(x)=x^3-4x^2+3x+4$ on the interval $[1,4]$.
So, with the equation $(b-a)\cdot f(c)=\int_1^4f(x)dx $, you get
$(4-1)\cdot f(c)=\int_1^4(x^3-4x^2+3x+4)dx$
Now my book says that this equals $3f(c)=\frac{57}{4}$.
I've been racking my brain and can't figure out how $\int_1^4(x^3-4x^2+3x+4)dx=\frac{57}{4}$
So how did the author evaluate that integral to get the answer?
 A: $\int_1^4(x^3-4x^2+3x+4)dx=\int_1^4x^3dx -4\int_1^4x^2dx +3\int_1^4xdx +4\int_1^4dx=:I$
$$\begin{align*}I &= \left.\left((1/4)x^4 -(4/3)x^3+(3/2)x^2 +4x\right)  
\right|_1^4\\ &=
(1/4)(4^4-1^4) -(4/3)(4^3-1^3)+(3/2)(4^2-1^2) +4(4-1)   
\\ &= (1/4)(255)-(4/3)(63)+(3/2)(15)+4(3)\\ &= 14.25\end{align*}$$
A: We have the following string of equalities:
$$\begin{align*}\int_1^4(x^3-4x^2+3x+4)dx &= \left.\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2}+4x\right|_{x=1}^4\\
&= \left(\frac{4^4}{4}-\frac{4\cdot4^3}{3}+\frac{3\cdot4^2}{2}+4\cdot4\right)-\left(\frac{1^4}{4}-\frac{4\cdot1^3}{3}+\frac{3\cdot1^2}{2}+4\cdot1\right)\\
&= \left(64-\frac{256}{3}+24+16\right)-\left(\frac{1}{4}-\frac{4}{3}+\frac{3}{2}+4\right)\\
&= \left(\frac{56}{3}\right)-\left(\frac{53}{12}\right)\\
&=\frac{57}{4}.\end{align*}$$
A: You've already seen that $$\int f(x)\,dx=\frac{x^4}4-\frac{4x^3}3+\frac{3x^2}2+4x+C.$$ As a corollary to the Fundamental Theorem of Calculus, if $F:[a,b]\to\Bbb R$ with $F(x)$ differentiable on $(a,b)$ and $F'(x)=f(x)$ on $(a,b)$ for some continuous $f:[a,b]\to\Bbb R$, then $$\int_a^bf(x)\,dx=F(b)-F(a).$$ In particular, letting $F(x)=\cfrac{x^4}4-\cfrac{4x^3}3+\cfrac{3x^2}2+4x+C$, with $a=1$ and $b=4$, we have $$\int_1^4f(x)\,dx=F(4)-F(1).$$ Since $F(4)=\cfrac{56}3+C$ and $F(1)=\cfrac{53}{12}+C$ (after some calculation), we get $$\int_1^4f(x)\,dx=\frac{171}{12}=\frac{57}4,$$ as desired. (Note that the "$+C$" bit didn't end up mattering.)
