Equations - AP Calculus AB

An definite integral can be solved by finiding the indefinite integral equation and plugging taking the difference of the values at the top and bottom numbers.

so the indefinite integral is

Plugging in the upper and lower limits gives

To integrate this expression, concentrate on each term separately. First, let's look at . The coefficient in the front (3) can be left out while integrating the x term. When integrating , you first raise the exponent by 1 and then put that result on the denominator as well, yielding .

Then, multiply by that by the coefficient that you left out in the beginning to get . Do the same for each term.

Integrating gives you .

Integrating gives you , and integrating x gives you 1. Then, string these all together (remembering appropriate negative signs!) to give you

.

Since this is an indefinite integral (there's not a particular solution), don't forget to add "C" to the end to give you a final answer of

.

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

The following for the second term:

And the following for the third term:

We can combine these terms and add our "C" to get the final answer:

To integrate this expression, you must use "u" substitution. The expression you assign to "u" is usually the expression with the higher exponent; in this case, . Since , the next step is to find du so that you can fully substitute everything.

The derivative of our u expression is

, or .

Now we can fully plug into our integral expression so we can integrate. You can rewrite the expression so it looks like this:

.

Everything in the u terms fully replaced everything in the original expression. Now, we can integrate. When we integrate , it becomes .

Since it's an indefinite expression, we must add +C at the end. The last step is to sub back in the expression that "u" represents.

Therefore, our final answer is

.

This integral requires "u" substitution. First assign an expression to u. In this case, . The next step is to find the derivative of u. Therefore, . Since there is no 2 in the original function, you must offset that by putting it on the "du" side: .

Now you can plug in everything so that the substitution is complete:

.

It might be helpful to put the outside the integral sign and to bring the u into the numerator:

.

Now, you can integrate the u term. Raise the exponent by 1 and also put that result on the denominator: . Then multiply that by the , which gives you . Now, sub back in the expression that u was representing: . Now, evaluate that expression from . When you plug in , you get . Plugging in yields . Now subtract those results to give you .

A good first step here is to simplify the expression. Since there's only one term on the bottom, you can make three separate terms out of the function. Therefore, rewriting it would look like:

and further simplifying it would look like this: .

Then, integrate each term. To integrate 2x, you must first raise the exponent by 1 and then put that result on the denominator, and then multiply by the coefficient 2: .

Anytime you integrate a constant, it becomes that constant multiplied by x; therefore 5x in this case. Integrating is a little trickier because of the x on the denominator. But whenever we have a single x on the denominator, we know that "ln" is involved and then we just multiply that by the coefficient: .

Also, since this is an indefinite integral, there must be a "+C" tacked on to the end.

Therefore, our final answer is:

.

To evaluate the integral, use the inverse power rule

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for the missing constants.

Reducing the quadratic term we arrive at our final answer.

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same umber. Don't forget to add a "plus C" to account for missing constants.

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem we get the following for the first term:

And for the second term:

We can combine these terms and add our "C" to get the final answer:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

And the following for the second term:

We can combine these terms and add our "C" to get the final answer:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

The following for the second term:

And the following for the third term:

We can combine these terms and add our "C" to get the final answer:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

And the following for the second term:

We can combine these terms and add our "C" to get the final answer:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

And the following for the second term:

We can combine these terms and add out "C" to get the final answer:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

And the following for the second term:

We can combine these terms and add our "C" to get the final answer:

An integral is the opposite of a derivative. For the general integration rules for this problem we need to know that the integral of is .

The integral of sine is negative cosine and the integral of is itself. We must also use the chain rule for sin and for the exponential function which states that the derivative of is .

The integral of is

.

The integral of is . The integral of is.

We must also add a C as the integral is indefinite.

The final answer is

.

An integral is the opposite of a derivative. The rule for integrating is .

This is enough information to find the integral of this function.

The integral can be found to be

The C is the constant of integration for indefinite integrals.

An integral is the opposite of a derivative. The integration rules needed for this problem are as follows.

The integral of is and the integral of cosine is sine.

This makes the integral

.

We must calculate this integral at the upper limit and subtract its value at the lower limit.

At the upper limit of we can calculate

At the lower limit of 0

so the final answer is

.

We know that the definite integral of acceleration between 0 and 4 seconds is the velocity. We know that the acceleration due to gravity is . We also know that the inititial velocity is . This means that when . Therefore, the velocity function can be written as

Here we want to integrate slices of the city with approximately uniform density. From the north side, we can divide the city into rectangles of length and width . Each rectangle has a population density of approximately , where is the distance from the north side of the rectangle. Thus, the area of each rectangle is and the total population of the rectangle is approximately people. We sum up all of the rectangles on the interval and take the limit as the number of rectangles on the interval approaches infinity. This limit takes the width of each rectangle and makes it infinitely small, making the density not just approximately uniform, but approaching perfectly uniform. This limit of a sum is one definition of the definite integral.

where and .