# Calculus 2 : Definite Integrals

## Example Questions

### Example Question #241 : Definite Integrals

Using the property of functions, solve the definite integral.

Explanation:

The first thing we must notice is that this is an odd function Given the properties of odd functions we know that it is a reflection over the origin.

This means that the same amount is above the axis as is below.

This means that the overall area is zero.

### Example Question #2355 : Calculus Ii

Solve the definite integral.

Explanation:

First we see that there is no special cases, so we can just use the power rule to integrate.

We add one to each power and then divide by the power so we end up with

.

After that we plug in our top bound and find our value.

Then we plug in our bottom bound and find our value. We take these two values and subtract the top value from the bottom value.

### Example Question #2356 : Calculus Ii

Solve the definite integral.

Explanation:

It is important to recognize this as it is a common and important integral, it is the natural log of x or more appropriately  . What we should look for when seeing this is 1/x or some variation of this.

When the integrals get harder and we start to do u-substitution, we will look for the derivative over the original.

But since in this problem, we know the integral will be , we just plug in the top value and get 1.

Plug in the bottom value and get 0.

Then subtract the two values to get 1.

### Example Question #2357 : Calculus Ii

Solve the definite integral.

Explanation:

This is another important integral to remember and that is e^x is always e^x.

Whether we integrate or derive, e^x will stay the same.

So it becomes important to just plug in out top bound, we get 5.

Plug in our bottom bound and we get 1.

So our answer is 5-1 which is 4.

### Example Question #2358 : Calculus Ii

Solve the definite integral.

Explanation:

For this problem we will use u-substitution. U-substitution can be thought of the reverse chain rule. We see we have a function we would have used the chain rule on and then we back track it. Here we can see that we have  and the neighbor down the block . This helps identify what the u for our substitution will be. so we take the stuff we have inside the parenthesis and we set that equal to u. So..

We derive that so we can find our dx.

Solve for dx.

Now we subsitute it back in.

Simplify.

Integrate.

Plug in original.

Plug in bounds.
512-1

Solve.

511

### Example Question #2359 : Calculus Ii

Find the definite integral.

Explanation:

First things is first, we identify what our u is. In this case, we want to look at our denominator. We see that we have  and in the numerator we have . We can see they are next door neighbors. So our u is...

Next we derive.

Solve for dx.

Substitute back in.

Simplify.

Integrate.

Plug in the original.

Plug in top bound and bottom pound.

### Example Question #2360 : Calculus Ii

Solve the definite integral using integration by parts.

Explanation:

First we must define what equation we will use to solve this integral. The equation to solve integration by parts is.

So we must define what our u and dv are. U is generally the simpler of the two terms or the natural log. Dv is generally the more complicated of the two. We will take the derivative of U and find the integral of dv.

So

Derive.

and

Integrate.

Now that we have all the parts, it is just plug and work out.

Take the second part and integrate.

Plug in top bound and bottom bound.

### Example Question #2361 : Calculus Ii

Solve the integral using integration by parts.

Explanation:

First we identify what are our u and dv are and then we derive u and take the integral of dv.

Derive.

Now the dv is a little tricky, because it looks like the only term there is lnx. We have to keep in mind that there is always a 1 attached to the front as a coefficient.

Integrate.

Now we plug into our by parts equation.

Simplify the second part.

Integrate the second part

Plug in the bounds.

### Example Question #2362 : Calculus Ii

Solve the integral by using integration by parts.

Explanation:

First things first. We must identify what our u and dv are. U is generally the simpler of the two and since 56x only has a coefficient we will go with that one. Dv will then be . We will need to take the derivative of u and integrate dv and then plug into the by parts equation.

Derive.

and

Integrate.

Plug into our by parts equation.

Integrate the second part.

Plug in the top value and the bottom value and subtract.

769