# Calculus 3 : Integration

## Example Questions

### Example Question #11 : Integration

A particle's velocity in two dimensions is described by the functions:

If the particle has an initial position of , what will its position be at time ?

Explanation:

Position can be found by integrating velocity with respect to time:

For velocities:

The position functions are:

These constants of integration can be found by using the given initial conditions:

Which gives the definite integrals:

### Example Question #12 : Integration

Calculate antiderivative of

Explanation:

We can calculate the antiderivative

by using -substitution. We set , so we get , so the integral becomes

So we just plug  to get

### Example Question #111 : Calculus Review

Calculate antiderivative of

Explanation:

We can calculate the antiderivative

by using the power rule for antiderivatives:

In this case , so we have

.

### Example Question #14 : Integration

The physical interpretation of the integral of a function , denoted by , is what?

No physical significance.

The average value of the tangent lines on .

The points of inflection of the function .

Area under the function .

Area under the function .

Explanation:

By definition, the integral of a fucntion  is the summation of an infinite number of small areas, thus giving the total area of the function with respect to the axis of integration.

### Example Question #11 : Integration

Calculate the integral of the function  given below.

Explanation:

We have a seperable integral, meaning each term can be integrated independently, written as

.

The first term is a power-rule integral, while the second is a trigonometric intergral.  One should recall

and

.

### Example Question #161 : How To Find Position

If the acceleration function of an object is , what is the position of the object at ? Assume the initial velocity and position is zero.

Explanation:

To find the position function from the acceleration function, integrate  twice.

When integrating, remember to increase the exponent of the variable by one and then divide the term by the new exponent. Do this for each term.

Solve for .

### Example Question #15 : Integration

Evaluate .

Explanation:

We can evaluate this integral using -substitution.

Let , then , hence we have

(Don't forget to change the bounds of integration)

.

### Example Question #12 : Integration

Evaluate , where is any constant.

Explanation:

Since is a constant, so is , therefore we can factor it out of the integral.

### Example Question #18 : Integration

Calculate .

Explanation:

This integral can be found using u-substitution.  Consider

.  This means we can rewrite our integral as

, by definition of the integral of an exponential.

### Example Question #17 : Integration

Calculate

Can not be determined.

Explanation:

This integral can be done using integration by parts.  Consider

.

Choose  and .

Using the definition of integration by parts,

,

.