Fundamental Theorems of Calculus

Fundamental Theorems of Calculus

Assessment

Interactive Video

Mathematics

11th - 12th Grade

Hard

Created by

Thomas White

FREE Resource

The video tutorial covers the second fundamental theorem of calculus, explaining how functions can be defined by integrals with variable bounds. It demonstrates calculating areas under curves using geometry and finding minimum and maximum values of functions. The tutorial also covers integrating and differentiating functions using the theorem, with practice problems and examples to reinforce learning.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main topic of today's lesson?

The First Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus

The Intermediate Value Theorem

The Mean Value Theorem

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is a function defined by an integral in this lesson?

By using a constant as the lower bound

By using a constant as the upper bound

By using a variable as either the lower or upper bound

By using a variable as both bounds

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What happens when the lower and upper bounds of an integral are the same?

The integral is undefined

The integral equals 0

The integral equals the function value

The integral equals 1

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the value of the integral from 0 to 1 of f(t) if the area under the curve is 1?

3

2

0

1

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you determine the minimum value of a function defined by an integral?

By finding where the derivative is zero

By evaluating the function at the endpoints

By finding the maximum area under the curve

By setting the integral equal to zero

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the maximum value of the function in the example provided?

At x = 2

At x = 0

At x = 4

At x = 1.5

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the result of taking the derivative of an integral?

The integral itself

Zero

A constant

The original function

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