Mean Value Theorem for Integrals

The function is differentiable at every point in the interval.

There exists a point where the area under the curve equals the area of a rectangle.

The function has a minimum value at the endpoint.

The function has a maximum value at the midpoint.

c = 2

c = 3

c = 4

c = 2.309

The derivative of the function at c.

The minimum value of the function.

The average value of the function over the interval.

The maximum value of the function.

c = 3

c = 6

c = 4.694

c = 5

Find the derivative of the function.

Calculate the definite integral over the interval.

Determine the endpoints of the interval.

Find the maximum value of the function.

c = 4

c = 6

c = 8

c = 10

It is the average of the function values at the endpoints.

It is the midpoint of the interval.

It is the maximum value of the function.

It is the minimum value of the function.

If the function is linear, c is always less than the midpoint.

If the function is concave up, c is less than the midpoint.

If the function is concave up, c is greater than the midpoint.

If the function is concave down, c is greater than the midpoint.

c is greater than the midpoint.

c is at the endpoint.

c is equal to the midpoint.

c is less than the midpoint.

It is the point where the function is not defined.

It is the point where the area under the curve equals the area of the rectangle.

It is where the function has its maximum value.

It is where the function has its minimum value.