Integration and Accumulation of Change Flashcards

What is the value of ∫ from 0 to 2 of (3x² − 2x) dx?

Answer: 4

An antiderivative is x³ − x². Evaluating from 0 to 2 gives (8 − 4) − 0 = 4.

If G(x) = ∫ from 1 to x² of sin(t) dt, what is G′(x)?

Answer: 2x·sin(x²)

By the Fundamental Theorem of Calculus with the chain rule, G′(x) = sin(x²)·d/dx(x²) = 2x·sin(x²).

Which of the following is equal to ∫ 2x·cos(x²) dx?

Answer: sin(x²) + C

Let u = x², so du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x²) + C.

A left Riemann sum with 4 subintervals of equal width is used to approximate ∫ from 0 to 4 of x² dx. What is the value of the sum?

Answer: 14

Each width is 1 and the left endpoints are 0, 1, 2, 3. The sum is 1·(0 + 1 + 4 + 9) = 14.

What is the value of ∫ from 0 to 1 of x·e^(x²) dx?

Answer: (e − 1)/2

Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫ from 0 to 1 of e^u du = (e − 1)/2.

What is the average value of f(x) = x² on the interval [0, 3]?

Answer: 3

The average value is (1/(3−0))·∫ from 0 to 3 of x² dx = (1/3)(9) = 3.

Water flows into a tank at a rate of r(t) = 2t + 1 gallons per minute. If the tank holds 5 gallons at t = 0, how many gallons are in the tank at t = 3 minutes?

Answer: 17

The amount is 5 + ∫ from 0 to 3 of (2t + 1) dt = 5 + [t² + t] from 0 to 3 = 5 + 12 = 17 gallons.

If ∫ from 2 to 6 of f(x) dx = 10 and ∫ from 2 to 4 of f(x) dx = 3, what is ∫ from 6 to 4 of f(x) dx?

Answer: −7

By additivity, ∫ from 4 to 6 of f = 10 − 3 = 7. Reversing the limits changes the sign, so ∫ from 6 to 4 of f = −7.

What is the value of ∫ from 0 to 1 of x/(x² + 1) dx?

Answer: (1/2)·ln 2

Let u = x² + 1, so du = 2x dx. The integral becomes (1/2)∫ from 1 to 2 of (1/u) du = (1/2)·ln 2.

Let F(x) = ∫ from 0 to x of (t² − 4t + 3) dt. At which value of x does F have a local maximum?

Answer: x = 1

F′(x) = x² − 4x + 3 = (x − 1)(x − 3). F′ changes from positive to negative at x = 1, so F has a local maximum there.