Fundamental Theorem of Calculus: Definite Intervals - AP Calculus BC
- Calculate the integral of from 1 to 4.
- . Same as , antiderivative is .
- Calculate the integral of from 1 to 2.
- . Antiderivative is , then .
- Find the integral of from 0 to 1.
- . Evaluate at bounds 1 and 0.
- What is the antiderivative of ?
- . Power rule integration: increase exponent, divide by new exponent.
- Find the integral of from 0 to 2.
- 4. Antiderivative is , evaluate at bounds.
- Identify the antiderivative given .
- . The antiderivative of is plus a constant.
- Determine the integral of from 0 to 1.
- . Antiderivative is , then .
- Find the integral of from 0 to 2.
- 16. Evaluate at bounds 2 and 0.
- Evaluate the integral of from 0 to 2.
- 4. Antiderivative is , then .
- Evaluate the integral of from 1 to 2.
- . Antiderivative of is , then .
- What does the Fundamental Theorem of Calculus connect?
- Differentiation and integration. Links the inverse operations of calculus.
- What is the integral of from 0 to 1?
- . Antiderivative is , evaluate at bounds.
- What is the definite integral of from 0 to 3?
- 15. Integral of constant function equals constant times interval length.
- Evaluate the integral of from 0 to 1.
- 0. Evaluate at bounds 1 and 0.
- Determine the integral of from 0 to 1.
- . Antiderivative of is , then .
- Calculate the integral of from 1 to .
- 1. Antiderivative of is , then .
- What is the integral of from to ?
- . Antiderivative of 1 is , difference of bounds gives length.
- What is the integral of from 1 to 3?
- 16. Antiderivative is , then .
- Find the integral of from 1 to 2.
- . Evaluate at bounds.
- Find the value of the integral of from 0 to 3.
- . Evaluate at bounds 3 and 0.