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.

The integral has a group raised to a power. This usually follows the basic integral form , where equals the group. 

Before we try to apply this, we should move the variables in the denominator up to the top. For our integral,, we can move the from the bottom up to the top by writing it with a negative exponent. This gives us

Now we try u-substitution, with the group as . Then we'll find by differentiating .

Our almost matches the from our integral, except it has a coefficient of , which is nowhere in our integral. Fortunately, is a constant that can be moved to the other side of the equation.

Now we can replace all the "x-stuff" with "u-stuff".

Notice that the integral bounds, and are changed to and . This is because the 1 and 8 are x values, but we are switching to u variables, and thus u values. This can be handled by "UN-substituting" back to x variables after integrating, or by finding the equivalent u values and plugging those in later. In this explanation we will stick with u variables and find the equivalent u values for the bounds.

To find the equivalent u bounds, we take the u substitution we used earlier, , and plug in each of the original bounds, and , then simplify for the equivalent u value for each. 

For ,

Now for ,

Having found the u-bounds, we will now begin to integrate. First we will pull the factor outside the integral.

Now we use the power rule for integration, .

The can be pulled out of the result and multiplied by the existing coefficient, .

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

is the result of integrating with the upper bound plugged in.

is the result of integrating with the lower bound plugged in.

For us, , our upper bound is , and our lower bound is .

Applying this theorem, we get

This is the correct answer.