# Calculus 1 : How to find integral expressions

## Example Questions

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### Example Question #181 : Integral Expressions

Solve the following integral, where a and b are constants:

Explanation:

Keeping in mind that a and b are only constants, the integral is equal to

and was found using the following rule:

### Example Question #181 : How To Find Integral Expressions

Evaluate the following integral:

Explanation:

To evaluate the integral, we must make the following substitution:

The derivative was found using the following rule:

Rewrite the integral in terms of u and integrate:

The integral was found using the following rule:

Finally, replace u with our original term:

.

### Example Question #181 : Integral Expressions

In circuits with a resistor, the equation for voltage drop is given by:

, where  is voltage,  is charge, and  is resistance.

Write the equation as an integral expression for

Explanation:

Although this may seem really difficult, we only need to solve for

To solve for , integrate both sides:

### Example Question #1231 : Functions

Explanation:

When integrating, remember to add one to the exponent and then put that result on the denominator: . Now evaluate at 2, and then 0. Then subtract the two results. .

### Example Question #182 : How To Find Integral Expressions

Explanation:

The first step here is to chop this up into three separate terms and then simplify since we have only one denominator: . Then, integrate each term, remembering to add one to the exponent and then put that result on the denominator: . Simplify to get your answer: . Remember to add C because it is an indefinite integral.

### Example Question #186 : Integral Expressions

Explanation:

First, chop this expression up into two terms: . Then, integrate each term, remembering that when there is a single x on a denominator, the integral is . Therefore, the integration is: . Remember to add C because it is an indefinite integral.

### Example Question #187 : Integral Expressions

Explanation:

First, integrate each term separately. Remember, when integrating, raise the exponent by one and then also put that result on the denominator: . Then evaluate at 3 and then 0. Subtract two results to get: .