### All Calculus 2 Resources

## Example Questions

### Example Question #1 : Area Under A Curve

Find the area under the curve from to .

**Possible Answers:**

**Correct answer:**

The area under a curve f(x) between two x-values corresponds to the integral

.

In this case, , and we can find by the fundamental theorem of calculus that...

evaluated from to

### Example Question #60 : Integral Applications

Find the area between and , such that and

**Possible Answers:**

**Correct answer:**

The first step to find the area between two curves is to set them equal to each other and find their intersection points.

The next step is to find which curve is "on the top" and which is "on the bottom" inside the interval . We find that is on the top and is on the bottom.

Now, to find the area between the two curves, we take the integral from to of the top curve minus the bottom curve.

evaluated from to

### Example Question #61 : Integral Applications

Find the total area under the curve from

**Possible Answers:**

**Correct answer:**

The definite integral of f(x) over the interval [0, 5] is the area under the curve.

A substitution makes this integral easier to evaluate. Let . Then . We can also change the limits of integration so that they are in terms of u. When x=5, . When x=0, . Making these substitutions, the integral becomes

### Example Question #62 : Integral Applications

What is the area under the curve over the interval ?

**Possible Answers:**

**Correct answer:**

The area under a curve over the interval is .

In this example, this leads to the definite integral

.

A substitution makes the antiderivative of this function more obvious.

Let

.

We can also convert the limits of integration to be in terms of to simplify evaluation. When , and when .

Making these substitutions results in

.

Recall that in general , so evaluating the integral leads to

### Example Question #63 : Integral Applications

Find the following indefinite integral:

**Possible Answers:**

**Correct answer:**

In order to find the anti-derivative, lets first factor the denominator.

Now we use partial-fraction decomposition to get this into a form we know how to take the anti-derivative.

Now we need to find the values of A and B. First lets write out the equation.

Now we can get our system of equations in order to find A and B.

From equation 1, we get

Now we can substitute this into equation 2.

Now we substitue B back into equation 1.

Now we can put the values of A, and B into our problem.

Now we can simply take the anti-derivative.

### Example Question #1981 : Calculus Ii

Evaluate:

**Possible Answers:**

**Correct answer:**

To evaluate the integral, we use inverse power law.

Remember that the inverse power law is

.

Lets apply this to our problem.

From here we plug in the bounds and subtract the lower bound function value from the upper bound function value.

### Example Question #2 : Area Under A Curve

Find the area under the curve of the function between and .

**Possible Answers:**

**Correct answer:**

Asking for the area under the curve means you are going to be doing an integral of the function provided. The bounds are given for you. The integral thus looks like this: .

When integrated, you get

evaluated at and .

At the expression evaluates to and at it evalutes to .

Subtracting the two gives you your final answer of .

### Example Question #61 : Integral Applications

Determine the area under the curve of

.

**Possible Answers:**

**Correct answer:**

For this particular function we will need to preform a "u-substitution".

In our case let

which will make .

Now we will substitute these into our integral to get the following.

if

Then we integrate using the power rule which states,

Now plug back in the original variable and then subtract the function values at the bounds.

### Example Question #1 : Area Under A Curve

Find the area under the curve of the following function from to :

**Possible Answers:**

**Correct answer:**

To find the area under the curve, we must integrate over the given bounds:

To integrate, we must do the following substitution:

The derivative was found using the following rule:

Now, rewrite the integral, and change the bounds in terms of u:

Note that during the rewriting, the bounds changed to the upper bound being -1 and the lower bound being 1, but the negative sign that came from the derivative of u allowed us to make the bounds as they are seen above.

Now perform the definite integration:

The integral was found using the following rule:

and the definite integration was finished by plugging in the upper bound into the resulting function, plugging the lower bound into the resulting function, and subtracting the two, as seen above.

### Example Question #68 : Integral Applications

Find the area underneath the curve of the function

on the interval

**Possible Answers:**

square units

square units

square units

square units

**Correct answer:**

square units

To find the area underneath the curve of the function on the interval , we find the definite integral

Because the function in this problem is always positive on the interval, or

on the interval the area underneath the curve can be found using the definite integral

and rewriting the function the definite integral is

Using the inverse power rule which states

we find the definite integral to be

And by the corollary of the Fundamental Theorem of Calculus, the definite integral equals

As such, the area is square units