Writing Equations - AP Calculus AB

can be written as with slope of and y-intercept of 1. So the inverse function would be or,

implies implies

To evaluate higher order derivative of trigonometric functions like the tangent function, we must become familiar with all of the other derivation rules; in this case we will use the Chain Rule and the Product Rule.

We begin with the identity that

Since the first derivative has now turned into a composite function in the form

where and

and and

we have that the second derivative, utilizing the Chain Rule, is

To obtain the third derivative, we must derive the second derivative via the Product Rule, since there is the multiplication of u=2\[sec(x)\]^2 and v=tan(x).

Recall that the Product Rule states that for a function, f(x)=uv:

As a result we have that

To solve, simply differentiate .

Remember to use the power rule.

Recall the power rule:

Apply this to our situation to get

To solve, simply differentiate and don't forget to indicate the differentiation by the "apostrophe" between the function and dependent variable. 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.