The way to find a dot product is to multiply the first number of one vector with the first number in the other. Then you do the same with the second and third numbers in each vector. Then you add up the three values and the answer is the dot product.

Recall the power rule for integration:

Break apart the integral into three integrals:

Integrate term by term,

Now add the constant of integration to the expression.

First, break apart the fraction into three fractions with each term having  in its denominator.

Then cancel the variables where you can and integrate term by term.  If you are left with any variables in the denominator then put them in the numerator with a negative sign.  

You get: 

Integrate these by the power rule,

We get:

Know the integrals and derivatives of trigonometric functions from memory, they occur very often. 

    and     

Now notice the pattern of 'going backwards' from taking derivatives to integrating:

   and    

We get, 

We have to use what is called direct substitution.  We let some quantity be represented by a new variable, and take the derivative of that.  Then we replace the quantity which was in the original equation, with this variable.  Commonly, mathematicians use the letter  for this substitution; nicknaming it a 'u-sub'.

Using direct substitution, Let   

Recall the power rule for differentiation:

then take the derivative of both sides of the equation,

.

You can replace    with    and now you can replace    with .

Now recall the power rule for integration: 

Finally, take the      fraction outside the integral and replace the old expression with the new variable. And then integrate and when you are done, replace the new variable back with the old quantity, as if you never even did it.

We have to use what is called direct substitution.  We let some quantity be represented by a new variable, and take the derivative of that.  Then we replace the quantity which was in the original equation, with this variable.  Commonly, mathematicians use the letter  for this substitution; nicknaming it a 'u-sub'.

 Begin by making the substitution and taking the derivative, and recall the power rule for differentiation.

And so,

Let   

.

Now we replace   with    and   with 

Recall the power rule for integration, 

Finally, pull out the    from the integrand and write the new integral and then solve.  After you solve the integral, put the original quantity back in.  That is replace  with what it originally equalled. 

First, graph the two functions in order to identify the boundaries of the region. You will find that they are .

Therefore, when you set up your integral, it will be from zero to one.

Next you set up your integral by subtracting the lower function from the upper function.

In this case, it would be .

This means your integral should look like 

.

Then solve using the power rule

:

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, it would be .

Therefore the integral would look like 

.

Then solve  using the rules for integrals of the trig functions:  

.

First, set up your integral, using the given bounds, by subtracting the lower function from the upper function.

In this case, in the given bounds, the integral will be 

.

Then solve  using the power rule

: 

First, set up your integral using the bounds found either by graphing or setting the two functions up to equal one another and solving.

Then subtract the lower function from the upper function.

In this case, in the given bounds, the integral will be 

.

Then solve  using the power rule

: