Use of the Fundamental Theorem to evaluate definite integrals - AP Calculus AB

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, 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 .

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 .

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).

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).

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.

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.

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 using the rule, then we apply the FTC

Now we just need to simplify,

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 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.