Evaluate the given definite integral:

Let's look at this integral as three separate parts, then we'll combine those parts to evaluate the answer.

First, the integral of  is just 

Second, The integral of any exponential term can be found by increasing the exponent by 1 and then dividing by the new number.

Third, the integral of cosine is simply positive sine.

Put all this together to get:

Now, we need to evaluate our answer by finding the difference between the limits of integration.

First, let's see what we get when we plug in 0

Next, let's use 5

Next, find the difference of F(5)-F(0)

So our answer is: -2388.89.

Evaluate the following definite integral.

Lets begin by recalling that taking an integral is essentially the same as reversing the differentiation process.

So, when we integrate sine, we get negative cosine, and when we integrate cosine, we get sine. The coefficients will stay the same.

Now, we have integrated, but we still need to evaluate our integral. To do so, we need to find the difference between  and .

Now, find the difference between our two values:

So, our answer must be 7.

Derivatives and anti derivatives annihilate each other.  Therefore, the derivative of an anti derivative is simply the function in the integral with the limits substituted in multiplied by the derivative of each limit.  Also, be careful that the units will match the outermost units (which comes from the derivative).

We will use the Fundamental Theorem of Calculus

First we find the anti derivative using the rule, then we apply the FTC

Now we just need to simplify,

Here we use the Fundamental Theorem of Calculus

and the rule

First we find the anti-derivative using the rule above (no constant is needed because we are dealing with definite integrals)

Then we use the FTC by plunging in 3 and subtracting from plugging in 0

We will use the Fundamental Theorem of Calculus 

and the rule

First we find the anti derivative

And then we evaluate, (upper minus lower)

(Remembering your logarithm rules)

We will use the Fundamental Theorem of Calculus

First we find the anti derivative

Then we evaluate it (upper minus lower)

Find the area of the region given by the following integral.

To evaluate this integral, recall the following rules:

Using these rules, we can find our integral. 

Clean it up a bit to get:

Alright, clean it up some more to find our final answer:

So, our area is 3737

To evaluate a definite integral, that is to say an integral with an upper bound and lower bound, we will use the 2nd Fundamental Theorem of Calculus, which states,

, where is the integral of .

We will first integrate just like with any integration.

We will integrate each term individually.

First, the integral of the cosine is the sine. Doing this, we get

The is just notation for "evaluated from to ". It is simply a reminder to apply the 2nd Fundamental Theorem of Calculus. The  is the from the theorem. Before moving to the second integral, we can apply this theorem.

So far, we have the following expression for the entire problem

The basic integral form for the remaining integral is

Set , since that is what is inside our sine. The derivative of this is

.

This perfectly accounts for the  in in front of the sine. Thus we can immediatly replace the entire integral with its basic integral form's result. 

Now we apply the 2nd Fundamental Theorem of Calculus again.

Plug in . Plug in and subtract.

This is our answer.

To evaluate this definite integral, we could try to fit it to one of the basic integral forms. The problem is that it doesn't match any of them. When this happens, we have two options: (1) Use algebra to force it to match a basic integral form, or (2) use algebra to break it into multiple terms that we can evaluate separately.  In this case, we will use option (2).

The two factors inside the integral can easily be multiplied together to get a polynomial. Doing this results in the following:

Combining like terms, we get

To integrate this polynomial, we integrate each terms separately. 

To evaluate the first three of these integrals, we will first pull their constant coefficients outside their integral.

The first three integral now match the power rule for integration. Recall that this rule states , which effectively says add 1 to the exponent to get the new exponent, and then multiply by the reciprocal of that new exponent. Applying this to the first three integrals gives us

Simplifying, we get

We will wait until after integrating the 4th term to write the  part. We will apply this to all 4 terms at once later. 

The fourth integral will follow the constant integration form, . Simply stated, multiply the constant by and "evaluate from a to b". "Evaluate from a to b" refers to the steps after integrating, using the 2nd Fundamental Theorem of Calculus. We will do this for all 4 terms after we finish integrating the 4th term.

Lets integrate the 4th term.

Now that we have used up all the integral symbols, we will apply the 2nd fundamental Theorem of Calculus, which, simply stated, says plug in the upper bound, then plug in the lower bound. Finally take the upper bound version minus the lower bound version to get the answer.

This is where we write the . This notation means that the upper bound is 2 and the lower bound is -1. These numbers are the same numbers as on the integral symbol from the original question, .

Applying the 2nd Fundamental Theorem of Calculus as described earlier, we get

Now we simplify to get our answer.

This is the correct answer.