Integration and Accumulation of Change Flashcards

What is the value of ∫ from 0 to 1 of x·eˣ dx?

Answer: 1

Integration by parts with u = x and dv = eˣ dx gives x·eˣ − eˣ. Evaluating from 0 to 1 gives (e − e) − (0 − 1) = 1.

Which expression equals ∫ 1/((x − 1)(x + 2)) dx?

Answer: (1/3)·ln|(x − 1)/(x + 2)| + C

Partial fractions give 1/((x−1)(x+2)) = (1/3)/(x−1) − (1/3)/(x+2). Integrating gives (1/3)ln|x−1| − (1/3)ln|x+2| + C.

If F(x) = ∫ from 0 to x² of √(1 + t³) dt, what is F′(x)?

Answer: 2x·√(1 + x⁶)

By the Fundamental Theorem of Calculus with the chain rule, F′(x) = √(1 + (x²)³) · d/dx(x²) = 2x·√(1 + x⁶).

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

Answer: 14

The 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 ∞ of e^(−2x) dx?

Answer: 1/2

The integral is the limit as b→∞ of [−(1/2)e^(−2x)] from 0 to b. That is 0 − (−1/2) = 1/2.

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 du/u = (1/2)ln 2.

What is the average value of f(x) = 3x² on the interval [1, 3]?

Answer: 13

The average value is (1/(3 − 1))·∫ from 1 to 3 of 3x² dx = (1/2)(27 − 1) = 13.

A tank holds 10 gallons of water at time t = 0. Water flows in at a rate of R(t) = 12 − 3t gallons per minute for 0 ≤ t ≤ 6, with t in minutes. How many gallons are in the tank at t = 6?

Answer: 28

The net change is ∫ from 0 to 6 of (12 − 3t) dt = 72 − 54 = 18 gallons. Adding the initial 10 gives 28 gallons.

What is the value of ∫ from 1 to e of x·ln x dx?

Answer: (e² + 1)/4

By parts with u = ln x and dv = x dx, the antiderivative is (x²/2)ln x − x²/4. Evaluating from 1 to e gives (e²/2 − e²/4) − (−1/4) = (e² + 1)/4.

What is the value of ∫ from 0 to 3 of 1/(x² + 9) dx?

Answer: π/12

∫ 1/(x² + 9) dx = (1/3)arctan(x/3) + C. Evaluating from 0 to 3 gives (1/3)(π/4 − 0) = π/12.