Here we use the fundamental theorem of Calculus: 

Here we do not worry about adding a constant c because we are evaluating a definite integral.

We can integrate this without too much trouble

. Start

. Rewrite the power

. Integrate

. Evaluate

. Simplify

Note that we were not asked to evaluate , so you should not attempt to use part one of the Fundamental Theorem of Calculus. This would give us the incorrect answer of .

To solve the integral, we first have to know that the fundamental theorem of calculus is 

.  

Since  denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6.  

To find the anti-derivative, we have to know that in the integral,  is the same as .  

The anti-derivative of the function  is , so we must evaluate .  

According to rules of logarithms, when subtracting two logs is the same as taking the log of a fraction of those two values: 

.  

Then, we can simplify to a final answer of 

To solve the integral using the Fundamental Theorem, we must first take the anti-derivative of the function. The anti-derivative of  is . Since the limits of integration are 1 and 3, we must evaluate the anti-derivative at these two values.  

 denotes the anti-derivative.  

When we do this, 

 and .  

The next step is to find the difference between the values at each limit of integration, because the Fundamental Theorem states 

.  

Thus, we subtract  to get a final answer of .

To solve the integral, we first have to know that the fundamental theorem of calculus is 

.  

Since  denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 0 and 3.  

The anti-derivative of the function  is , so we must evaluate .  

When we plug 3 into the anti-derivative, the solution is , and when we plug 0 into the anti-derivative, the solution is 0.  

To find the final answer, we must take the difference of these two solutions, so the final answer is .

To solve the integral, we first have to know that the fundamental theorem of calculus is 

.  

Since  denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 0 and 2.  

The anti-derivative of the function 

is 

,

so we must evaluate .  

When we plug 3 into the anti-derivative, the solution is , and when we plug 0 into the anti-derivative, the solution is 0.  

To find the final answer, we must take the difference of these two solutions, so the final answer is .

First compute the indefinite integral: 

Note that the  is the derivative of . So proceed by defining a new variable: 

Now the the integral can be written in terms of 

Therefore: 

When we go to compute the indefinite integral the constant of integration  will be ignored since it will be subtracted out when we evaluate. 

We can precede by either going back to the original variable  and evaluate over the original limits of integration, or we can find new limits of integration corresponding to the new variable . Let's look at both equivalent methods: 

Solution 1)

  so the last term vanishes. The first term reduces to  since the tangent function is equal to . 

Solution 2) 

We could have also solved without converting back to the original variable. Instead, we could just change the limits of integration. Use the definition assigned to the variable , which was  and then use this to find which value  takes on when  (lower limit) and when  (upper limit). 

This is a Fundamental Theorem of Calculus problem.  Since a derivative and anti-derivative cancel each other out, we simply have to plug the limits into our function (with the outside variable).  Then, we multiply each by the derivative of the bound:

Using the Fundamental Theorem of Calculus, the derivative of an anti-derivative simply gives us the function with the limits plugged in multiplied by the derivative of the respective bounds:

In the last step, we made use of the following trigonometric identity:

Evaluate the following indefinite integral:

Recall that we can split subtraction and addition within integrals into separate integrals. This means that we can look at our problem in two steps.

Recall that we can integrate any exponential term by adding 1 to the exponent and dividing by the new exponent.

So,

Next, recall that the integral of sine is negative cosine. However, we already have a negative sine, so we should get positive cosine.

Now, we can combine our two halves to get our final answer. 

Notice that we only have one "c" because c is just a constant, not a variable.