# Homework Help: Apply the mean value theorem for integrals

1. Feb 1, 2012

### bobsmith76

1. The problem statement, all variables and given/known data

3. The attempt at a solution

My book is not explaining very well the steps at solving these problems. There is a step that I'm missing

step 1. find 1/(b-a) easy
step 2. find the antiderivative of 4 - x, easy, x^2/2
step 3. plug in what the result of the antiderivative is for b and a, subtract b from a
step 4. this is the step I'm missing. the books shows that I'm supposed to multiply 4 by b - a. why? what is this? I need to know what mathematical terminology this is so that i can do these problems in the future
step 5. subtract step 4 from step 3. why? what mathematical terminology is this.
step 6. multiply step 1 by step 5, I understand that.

I'm not yet able to read the notation and the book is not helping.

2. Feb 1, 2012

### HallsofIvy

Well, 4x- x^2/2

The anti-derivative of 4, with respect to x, is 4x. That is evaluated between the limits of integration, b and a. You can do it as either 4b- 4a or 4(b- a). They are equal by the "distributive law".

No, they are not subtracting step 4 from step 3. They are simply completing the calculation indicated in step 4. There are, however, different ways of doing that step. You could have calculated that 4(3)= 12, and $3^2/2= 9/2$ and subtract: 12- 9/2= 24/2- 9/2= 15/2.
Then multiply by the leading 1/3: (1/3)(15/2)= 5/2.

Or, using the distributive law, go ahead and multiply by the 1/3 first: (1/3)(4(3)- 3^2/2)= (1/3)(4(3))- (1/3)(3^2)/2. That 1/3 will cancel a 3 in eacy term: it is equal to 4- 3/2= 8/2- 3/2= 5/2 again.

The distributive law: a(b+ c)= ab+ ac.

3. Feb 1, 2012

### bobsmith76

Ok, thanks, I incorrectly calculated the antiderivative of 4 - x as (x^2)/2, it's (4x - x^2)/2.

4. Feb 1, 2012

### genericusrnme

All that's happened here is an integral
Here are some properties of integrals that you should know that will also help you understand your problem

Let a be a constant, f(x) be some function of x, and F(x) be it's anti-derivative

1. $\int (a \pm f(x))dx = \int a dx \pm \int f(x) dx$

2. $\int a f(x) dx = a \int f(x) dx$

3. $\int 1 dx = x$

4. $\int_a ^b f(x) dx = F(b) - F(a)$

Now I shall show you which of these properties we shall use in our problem;

$\frac{1}{b-a} \int_a ^b f(x) fx = \frac{1}{b-a} \int_a ^b (4-x)dx$

I shall use property 1. now

$\frac{1}{b-a} \int_a ^b (4-x)dx$ = \frac{1}{b-a}( \int_a ^b 4 dx - \int_a ^b x dx[/itex]

I shall now use property 2. on the first term

$\frac{1}{b-a}( \int_a ^b 4 dx - \int_a ^b x dx = \frac{1}{b-a} ( 4 \int _a ^b 1 dx - \int_a ^b x dx)$

Due to property 3, the anti-derivative of 1 is x (or in the terms of property 4. if f(x)=1 then F(x) = x)
And using property 4 to evaluate the definite integrals

$\frac{1}{b-a} ( 4 \int _a ^b 1 dx - \int_a ^b x dx) = \frac{1}{b-a} ( 4(b-a) -( \frac{b^2}{2} -\frac{a^2}{2} ))$

Substituting in our values for a and b we get

$\frac{1}{b-a} ( 4(b-a) -( \frac{b^2}{2} -\frac{a^2}{2} )) = \frac{1}{3} ( 4(3) -\frac{3^2}{2}) = 4 - \frac{3}{2} = \frac{5}{2}$

Has this helped at all?