Mean Value Theorem - AP Calculus AB

The area of this region is given by the following integral:

or

Taking the antiderivative gives

, evaluated from to .

, and

.

Thus, the area is given by:

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

int_${2}^{7}$$\frac{1}{sqrt{x+2}$}dx=2$\sqrt{x+2}$Big|_$2^7$=2

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by and in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve is .

The area of the triangle above the curve is 2. Therefore, the area bounded is .

The two curves intersect in between and , which can be found by solving the quadratic equation $12-x^2$=-x.

To solve for the area between curves, and , we use the formula $A=int^b$_a(f(x)-g(x))dx

For our problem:

$A=int^4$_${-3}(12-x^2$-(-x))dx

$A=12x-$\frac{x^3$$}{3}$+\frac{x^2$}{2}$ evaluated from to which yields .

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be such that . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be such that .

is given to be continuous. Also, if we set , we note that . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be such that .

Incidentally, it does follow from the given information that must have a zero on the interval , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on , then there exists at least one value such that . , a polynomial, is continuous and differentiable everywhere; setting , it follows from the MVT that there is such that

Evaluating and :

The expression for is equal to

,

the correct choice.

To find the mean value of a function over some interval , one mus use the formula: .

Plugging in

Simplifying

One must then use the inverse Sine function to find the value c:

The area of this region is given by the following integral:

or

Taking the antiderivative gives

, evaluated from to .

, and

.

Thus, the area is given by:

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

int_${2}^{7}$$\frac{1}{sqrt{x+2}$}dx=2$\sqrt{x+2}$Big|_$2^7$=2

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by and in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve is .

The area of the triangle above the curve is 2. Therefore, the area bounded is .

The two curves intersect in between and , which can be found by solving the quadratic equation $12-x^2$=-x.

To solve for the area between curves, and , we use the formula $A=int^b$_a(f(x)-g(x))dx

For our problem:

$A=int^4$_${-3}(12-x^2$-(-x))dx

$A=12x-$\frac{x^3$$}{3}$+\frac{x^2$}{2}$ evaluated from to which yields .

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be such that . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

By Rolle's Theorem, if is continuous on and differentiable on , and , then there must be such that .

is given to be continuous. Also, if we set , we note that . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be such that .

Incidentally, it does follow from the given information that must have a zero on the interval , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on , then there exists at least one value such that . , a polynomial, is continuous and differentiable everywhere; setting , it follows from the MVT that there is such that

Evaluating and :

The expression for is equal to

,

the correct choice.

To find the mean value of a function over some interval , one mus use the formula: .

Plugging in

Simplifying

One must then use the inverse Sine function to find the value c:

The two curves intersect in between and , which can be found by solving the quadratic equation $12-x^2$=-x.

To solve for the area between curves, and , we use the formula $A=int^b$_a(f(x)-g(x))dx

For our problem:

$A=int^4$_${-3}(12-x^2$-(-x))dx

$A=12x-$\frac{x^3$$}{3}$+\frac{x^2$}{2}$ evaluated from to which yields .

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

int_${2}^{7}$$\frac{1}{sqrt{x+2}$}dx=2$\sqrt{x+2}$Big|_$2^7$=2