- Example Questions
- Example Question #231 : Definite Integrals
- Example Question #232 : Definite Integrals
- Example Question #233 : Definite Integrals
- Example Question #234 : Definite Integrals
- Example Question #235 : Definite Integrals
- Example Question #236 : Definite Integrals
- Example Question #237 : Definite Integrals
- Example Question #238 : Definite Integrals
- Example Question #239 : Definite Integrals
- Example Question #240 : Definite Integrals
- All Calculus 2 Resources

### All Calculus 2 Resources

## Example Questions

### Example Question #231 : Definite Integrals

### Example Question #232 : Definite Integrals

### Example Question #233 : Definite Integrals

### Example Question #234 : Definite Integrals

### Example Question #235 : Definite Integrals

What is

For this problem, we need to first simplify the integral by splitting in two integrals.

Now, the first integral was given in the problem statement. Therefore, we just need to add that to the evaluation of the second integral.

Add the two parts together:

### Example Question #236 : Definite Integrals

Evaluate this integral.

Now let thus .

Also notice sin(0)=0 and .

So we have

evaluated between 0 and 1.

We have

### Example Question #237 : Definite Integrals

Determine the area bounded by and between on .

First identify the higher and lower function.

The higher function:

The lower function:

Since between and we must integrate.

To integrate take the difference of the lower function from the higher function.

Evaluating between zero and one results in,

### Example Question #238 : Definite Integrals

Evaluate

Evaluated between 0 and 1.

### Example Question #239 : Definite Integrals

Using geometry, Find the definite integral of the function.

When we look at the figure we are trying to integrate, we see that it is actually a quarter circle.

Using the rules we learned in geometry, we know that the area of a circle is . As such, we can take the radius and square it and then multiply it by . Our radius is found by taking the square root of the constant under the radical.

So now that we have the area of the whole circle, we need to divide it by 4.

### Example Question #240 : Definite Integrals

Using geometry, solve the definite integral.

When we look at the shape of the graph we notice that it is a triangle with a height of 3 and a base of 3.

So the area under the curve would be the same as the area of this triangle.

From geometry, from geometry we know that the area of a triangle is

.

So our answer would be

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### All Calculus 2 Resources