Set up, but do not evaluate, an integral expression for the following:

Find the area of the region bound by the lines , the y-axis, and h(t). 

To begin, we should find our limits of integration. In this case, we are told the region is bound by two lines and the y-axis. The two lines should be our lower and upper limits of integration, making our integral expression as follows.

Recall that our function always goes inside the integral, and that the lower limit goes at the bottom, while the upper limit goes at the top.

Don't be fooled into switching the limits of integration. We want the smaller number to be on the bottom.

Solving this integral requires a simple u-substitution. We need to find some way to turn this integral into the form of:

For which we know the result is simply:

Taking the original integral, and making the appropriate substitions will result in this:

Once this substition has been made, the integral must be adjusted for this change. 

Because  , then 

Making the appropriate substitutions results in:

This is a simple integral, for which we already know the result:

Now replacing the original value for u, results in the final expression:

Solving this integral requires a simple u-substitution. We need to find some way to turn this integral into the form of:

For which we know the result is simply:

Taking the original integral and making the appropriate subsitutions will result in this:

Once this substitution has been made, the integral must be adjusted for this change.

Because  , then 

Making the appropriate substitions results in:

This is a simple integral, for which we already know the result:

Now replacing the original value for u, results in the final expression:

Solving this integral requires a simple u-substitution. We need to find some to turn this integral into the form of:

 , Where a is a constant, and u is a differentiable function

We know the result of this integral is:

Taking the original integral and making the appropriate substitutions will result in this:

 ,

Since 5 is a constant term

Once this substitution has been made, the integral must be adjusted for this change.

Because  , then 

Making the appropriate substitutions results in:

Pulling the  outside of the integral results in:

Using the formula from above, this integral turns into:

Now replacing the original value for u, results in the final expression:

Solving this integral requires a simple u-substitution. We need to find some to turn this integral into the form of:

 , Where a is a constant, and u is a differentiable function.

We know the result of this integral is:

Taking the original integral and making the appropriate substitutions will result in this:

Since 2 is a constant term, we can pull it outside of the integral.

Once this substitution has been made, the integral must be adjusted for this change.

Because  , then 

Making the appropriate substitutions results in:

Pulling the  outside of the integral results in:

Using the formula above, this integral turns into:

Now replacing the original value for u, results in the final expression:

Solving this integral requires a simple u-substitution. We need to find some to turn this integral into the form of:

 , Where a is a constant, and u is a differentiable function.

We know the result of this integral is:

Notice, also that this is a definite integral and so, when making the u-substitution, the bounds must be accounted for.

Taking the original integral and making the appropriate substitutions will result in this:

Since 2 is a constant term, we can pull it outside the integral.

Because this is a definite integral we must adjust the bounds accordingly. The lower bound when the integral is in terms of x is 0, and the upper bound is 2. 

When we change the integral to be in terms of u, the bounds change like so:

 Lower Bound

 Upper Bound

Once this substitution has been made, the integral must be adjusted for this change.

Because  , then 

Making the appropriate substitutions results in:

Pulling the  outside of the integral results in:

We know that  , and using the formula from above, this integral turns into:

Now, plugging in the upper and lower bounds into u results in:

An definite integral can be solved by finiding the indefinite integral equation and plugging taking the difference of the values at the top and bottom numbers.

so the indefinite integral is 

Plugging in the upper and lower limits gives

An integral is the inverse operation of a derivative. The derivative of  is . The derivative of  is .

Therefore the

and 

The constant C must be included because the integral is indefinite. The final answer is

.

This type of integral cannot be solved by normal methods so we will use integration by parts. Integration by parts states

If  then  because the derivative of 

If  then  because the derivative of  is 

so the integral is equal to 

To integrate this expression, concentrate on each term separately. First, let's look at . The coefficient in the front (3) can be left out while integrating the x term. When integrating , you first raise the exponent by 1 and then put that result on the denominator as well, yielding .

Then, multiply by that by the coefficient that you left out in the beginning to get . Do the same for each term.

Integrating gives you .

Integrating gives you , and integrating x gives you 1. Then, string these all together (remembering appropriate negative signs!) to give you

.

Since this is an indefinite integral (there's not a particular solution), don't forget to add "C" to the end to give you a final answer of

.