Equations - AP Calculus AB

Note that:

Product Rule:

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

The derivative is then found using the product rule to be:

Notice how the chain rule needs to be utilized an additional time when taking the derivative of the term with respect to .

Now bring and terms to opposite sides of the equation:

Now rearraging variables gives :

For this problem, note that:

Product Rule

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

If we're looking for for the function then we'll begin by differentiating with respect to first:

Next, differentiate with respect to :

Now finally we'll differentiate with respect to ; remember to use the product rule:

To solve, simply find the first derivative and let . Thus,

Recall that area under the curve requires integration from either top minus bottom curve, or right minus left curve.

The top curve is:

The bottom curve is:

The interval indicates that the lower bound is 2 and the upper bound is 5.

Write the integral.

By normal exponent rules , and so we set up the definite integral

,

which integrates to:

Integrating by parts with

,

, ,

we get:

This problem requires you to evaluate an indefinite integral of the given function f’(x). Integrating each term of the function with respect to x, we simply divide each coefficient by (n+1), where n is the value of the exponent on x for that particular term, and then add 1 to the value of the exponent on each x. This gives us:

A correct answer must include a constant C, as the original function may have had a constant that is not reflected in the equation for its derivative.

We start by integrating the function in parentheses with respect to x, and we then subtract the evaluation of the lower limit from the evaluation of the upper limit:

This problem intends to demonstrate one possible application of calculus to the field of physics. An equation for work as given in physics is as follows:

Using this equation, we simply set up a definite integral with our bounds given by the interval of interest in the problem:

Joules

After evaluating our integral, we can see that the work required to push the box from x=2 meters to x=5 meters is 39.8 Joules.

Simply use an integral expression with the given x values as your bounds.

Don't forget to add the dx!

In order to solve this problem we must know that

In this case we have:

Our first step is to integrate:

We then arrive to our solution by plugging in our values

In order to solve this problem we must remember that:

In this case we have:

Our first step is to integrate:

We then arrive at our solution by plugging in our values

In order to solve this problem we must remember that:

In this case we have:

Our first step is to integrate:

We then arrive to our solution by plugging in our values

We are being asked to integrate the function.

To do this we need to remember the power rule of integrals,

Using this rule, we can evaluate the following integral:

In this problem we are being asked to integrate the function.

In order to do this we need to remember,

.

Since

then the integral,

.

This problem is asking us to integrate the function.

In order to do this we need to remember,

.

Therefore,

We are asked to set up an integral without solving it. We want to evaluate the function from 3 to 13, so 3 goes at the bottom of the integral and 13 goes at the top. No other options have this correct.

This is a tricky question.

If we chose , the region in the third quadrant of is negative area, since is no longer the top curve. Evaluating this integral will cancel out the negative area in the third quadrant with the positive area on the first quadrant. This integral will give zero area, which is incorrect.

The correct method is to split this interval into 2 separate integrations: One from and the other from .

From interval , the top curve is minus the bottom curve, , is .

From interval , the top curve is minus the bottom curve, , is .

Set up the integral.

To find the area under a curve, we usually want to use an integral. All of our answer choices are integrals though, so this doesn't narrow things down. Then next thing to look for are our limits of integration. In this case, we are on the interval \[360,540\], so look for an option that has the lower limit of integration as 360 and the higher limit as 540. Only one option has this:

For further clarification, to set up an integral, we want our function within the integral and our limits of integration on the integral sign itself. Once we have the correct integral set up, we can evaluate it to find the area under the curve.

The region bounded by the three functions represents a triangle. It will be easier to determine the bounded region by subtracting the left curve from the right curve, and using to integrate. In order to integrate in terms of , any equation in terms of must be rewritten in terms of .

Isolate .

Find the intersection of the lines and .

The correct bounds are from . Write the correct integral.