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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?

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When you evaluate that integral, what do you get? – mixedmath Aug 20 '12 at 17:33
I get $\frac{57}{4}$. The indefinite result is $\frac{x^4}{4} - \frac{4 x^3}{3} + \frac{3 x^2}{2} + 4 x$. – copper.hat Aug 20 '12 at 17:34
Well I used the Power Rule to get $\frac{x^4}{4}-4\cdot \frac{x^3}{3}+3\cdot \frac{x^2}{2}+4x$ but then I don't know what to do from there? – Olly Price Aug 20 '12 at 17:43
What do I sub into $x$ to get $\frac{57}{4}$? – Olly Price Aug 20 '12 at 17:43
It is a definite integral, so evaluate it with $x=4$ and subtract the result of evaluating at $x=1$. – copper.hat Aug 20 '12 at 17:46
up vote 1 down vote accepted

We have the following string of equalities: $$\begin{align*}\int_1^4(x^3-4x^2+3x+4)dx &= \left.\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2}+4x\right|_{x=1}^4\\ &= \left(\frac{4^4}{4}-\frac{4\cdot4^3}{3}+\frac{3\cdot4^2}{2}+4\cdot4\right)-\left(\frac{1^4}{4}-\frac{4\cdot1^3}{3}+\frac{3\cdot1^2}{2}+4\cdot1\right)\\ &= \left(64-\frac{256}{3}+24+16\right)-\left(\frac{1}{4}-\frac{4}{3}+\frac{3}{2}+4\right)\\ &= \left(\frac{56}{3}\right)-\left(\frac{53}{12}\right)\\ &=\frac{57}{4}.\end{align*}$$

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$\int_1^4(x^3-4x^2+3x+4)dx=\int_1^4x^3dx -4\int_1^4x^2dx +3\int_1^4xdx +4\int_1^4dx=:I$

$$\begin{align*}I &= \left.\left((1/4)x^4 -(4/3)x^3+(3/2)x^2 +4x\right) \right|_1^4\\ &= (1/4)(4^4-1^4) -(4/3)(4^3-1^3)+(3/2)(4^2-1^2) +4(4-1) \\ &= (1/4)(255)-(4/3)(63)+(3/2)(15)+4(3)\\ &= 14.25\end{align*}$$

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How did you come to the values of 255, 63, 15 and 3? – Olly Price Aug 20 '12 at 17:46
I've edited the post for you ;) – Zeta.Investigator Aug 20 '12 at 17:47
Aha, thank you! – Olly Price Aug 20 '12 at 17:47

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.)

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