Other Writing Equations - AP Calculus AB

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 2. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 6. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 3. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 100. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 4. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 12. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.

To find the area under a curve, you need to use an integral expression. The number at the base of your integral sign is your lower bound (the smaller x-value that defines one vertical edge of your area), which in this case is 5. The number at the top of your integral sign is your upper-bound (the larger x-value that defines the other vertical edge of your area), which in this case is 8. The equation that goes inside of your integral expression is the same equation that you were initially given, without the "y = ". Finally, you need to add "dx" after your equation in order to complete it.

To solve, you must first differentiate according to the power rule and then plug in 1. Thus,

Power rule:

Since for all values of x, the first derivative is 6, our answer is 6.

This problem is asking you to find the indefinite integral of the given function, f'(x). To do this, we need to work through the function one term at a time. For each term in the function, we perform two steps. First, we divide the coefficient of that term by (n+1), where "n" is the value of the exponent for that term's x-value. Second, we add 1 to the value of that term's x-value. We continue this process for each term in the function.

So, for , this would look like

which when integrated will equal

You must remember to include "C" in your final answer; this acts as a placeholder for any constants that may have been present in the original equation.

Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x."

Now we can look at the two remaining answer choices. Let y=f(x) and solve for x to find the inverse.

So f(x)=y=3x+1. Solve for x.

x=(y-1)/3

Therefore our answer is (y-1)/3.

The polar equation formula here is x2+y2=r2.

This is the formula for a cirlce, so we can eliminate the two "line" answers.

Plugging in r=-6 into the equation gives x2+y2=36, which describes the graph of a circle with radius 6.

First we must solve for the general derivative of f(x) = f(x) = x4 + 5x2 + 44x – 3.

f'(x) = 4x3 + 10x + 44

Now, the slope of the tangent line for f(15) is equal to f'(15):

f'(15) = 4(15)3 + 10 *15 + 44 = 13694.

To find the tangent line, we need at least one point on the line. To find this, we can use f(15) to get the y value of the point of tangency, which will suffice for our use:

f(15) = 154 + 5(15)2 + 44 * 15 – 3 = 50625 + 1125 + 660 – 3 = 52407

Now, using the point-slope form of the line, we get:

y - 52407 = 13694 * (x – 15)

Simplify:

y – 52407 = 13694x – 205410

y = 13694x – 153003

First, find the y coordinate of the function at

Thus, we have the point .

To find the slope of the tangent line, take the derivative and then plug in

Thus, our slope is .

Knowing that the general formula of a straight line is where is the slope and that slope is . Thus, . To find , plug in the coordinate and then solve for .

So now the equation of the tangent line becomes

First, foil the integral so it is easier to manage.

then perform the indefinite integral the normal way you would do

In this limit, we have as .

Because of this, we can use L'Hospital's rule. Differentiating the top and the bottom of the function, we get

.

If we evaluate at , we don't have anymore and we get

.

So the answer is 2.

Because we have a in our limit, we can use L'Hospital's Rule. Applying this, we get

.

Evaluating this as we get

Our first step is to apply implicit differentiation to the function to find y’. This derivative tells us the slope of our function and will therefore give us the slope of our tangent line at the given point:

Rearranging the equation, factoring out y’, and dividing, we find y’ is:

Before we can plug in the values of our given point to find the slope of the tangent line, we must first find its y value by plugging x=0 into the original function, which gives us:

So now that we know our given point is (0,3), we plug these values into our equation for y’, which gives us the slope of the tangent line at that point.

Now that we have our slope (m), an x coordinate (x1), and a y coordinate (y1), we can simply plug these values into the point-slope formula for a line:

This problem deals with the concept of optimization. We start by simply writing out the equations described in words by the problem:

The problem asks that we maximize the product of the two unknown numbers, C, but we must first solve our first equation for y. Then we substitute it into our second equation to ensure that it is in terms of x alone. We then take the derivative of the equation with respect to x, and set it equal to 0 to solve for the x value that corresponds to the critical point of the function:

Because we only know that this x value corresponds to a critical point, and not necessarily whether that point is a minimum or a maximum, we must check the value of the second derivative at this point to see whether the function is concave up or concave down:

Because the value of the second derivative at this point is negative, we know that the function is concave down, so our one and only critical point in this case occurs at a global maximum, verifying that the product of the two numbers is maximized when x=225. Now that we know the value of x, we substitute it into the equation for y to find our second unknown number:

To write this equation we need to read the question carefully. Since the slope is we know the slope is represented by .

Also since the y-intercept is , when is then .

Therefore plugging our values into the slope intercept form:

we get,

.

My friend starts off with dollars, and every day, she makes a net dollar. This is because she earns three dollars and spends one dollar.

So, we need an equation that reflects that she has *(number of days), which is given by

.

The first section of the equation to be resolved (in this case 1-4x) is the innermost function and the second section to solve is the outer function. So:

becuase g(x) must be performed first before plugging it into f(x).

The first section of the equation to be resolved (in this case 2x^3+5) is the innermost function and the second section to solve is the outer function.

So:

becuase g(x) must be performed first before plugging it into f(x).

The first section of the equation to be resolved (in this case pi * x) is the innermost function and the second section to solve is the outer function.

So:

becuase g(x) must be performed first before plugging it into f(x).