Difficulty: Introductory to Intermediate | Prerequisites: Elementary Integration core concepts (power rule, trig antiderivatives, FTC)
This document walks through 15 worked problems from an NYU Calculus II WebAssign set on elementary integration. The problems cover algebraic simplification, polynomial integrals, radical/fractional-exponent integrals, trigonometric integrals, and inverse-trig integrals. Each solution is broken into steps so you can see exactly where the algebra and the calculus happen. Use this alongside the Core Concepts notes.
Fifteen problems in four categories: algebra warm-ups, polynomial definite integrals, integrals with roots and fractional exponents, and trig/inverse-trig integrals. The recurring lesson is the same: simplify first, apply the power rule or the relevant formula, then evaluate at the bounds carefully.
These are not integrals yet. They test the algebra you will need before you can integrate.
Use FOIL: x·x + x·(-2) + 1·x + 1·(-2)
= x^2 - 2x + x - 2
= x^2 - x - 2
FOIL: 12r^2 - 24rs - 15rs + 30s^2
= 12r^2 - 39rs + 30s^2
Factor out 3: 3(4r^2 - 13rs + 10s^2)
Split the fraction: 6t^2/(3t) - t/(3t)
= 2t - 1/3
This kind of division is exactly what you do to simplify an integrand before applying the power rule.
Same principle as Problem 3. The pattern is: when a polynomial is divided by a monomial, split into separate fractions and simplify each term. You will use this technique repeatedly in the integration problems that follow.
Antiderivative: x^3/3 + x^2 - 5x
Evaluate at upper bound (3): 27/3 + 9 - 15 = 9 + 9 - 15 = 3
Evaluate at lower bound (1): 1/3 + 1 - 5 = 1/3 - 4 = -11/3
Subtract: 3 - (-11/3) = 3 + 11/3 = 9/3 + 11/3 = 20/3
Expand first: (u + 3)(u - 4) = u^2 - u - 12
Antiderivative: u^3/3 - u^2/2 - 12u
Evaluate at 1: 1/3 - 1/2 - 12 = 2/6 - 3/6 - 72/6 = -73/6
Evaluate at 0: 0
Result: -73/6
Note the negative result. A definite integral can be negative when the function lies below the x-axis over the interval.
Expand: 6x + x^4
Antiderivative: 3x^2 + x^5/5
Evaluate at 2: 3(4) + 32/5 = 12 + 32/5 = 60/5 + 32/5 = 92/5
Evaluate at 0: 0
Result: 92/5
Antiderivative: x^4/4
Evaluate at 2: 16/4 = 4
Evaluate at -1: (-1)^4/4 = 1/4
Subtract: 4 - 1/4 = 15/4
Watch the sign: (-1)^4 = 1, not -1. Even powers of negative numbers are positive.
Rewrite: sqrt(x) = x^(1/2)
Antiderivative: x^(3/2) / (3/2) = 2x^(3/2) / 3
Evaluate at 9: 2(9)^(3/2) / 3 = 2(27)/3 = 54/3 = 18
Evaluate at 4: 2(4)^(3/2) / 3 = 2(8)/3 = 16/3
Subtract: 18 - 16/3 = 54/3 - 16/3 = 38/3
Key step: (9)^(3/2) means (sqrt(9))^3 = 3^3 = 27. Similarly (4)^(3/2) = (sqrt(4))^3 = 2^3 = 8.
Distribute: 7t^(1/2) + t^(3/2)
Antiderivative: 7 · 2t^(3/2)/3 + 2t^(5/2)/5 = 14t^(3/2)/3 + 2t^(5/2)/5
Evaluate at 4: 14(8)/3 + 2(32)/5 = 112/3 + 64/5 = 560/15 + 192/15 = 752/15
Evaluate at 0: 0
Result: 752/15
The crucial move is distributing the sqrt(t) across the sum before integrating.
Split and rewrite: 4/sqrt(x) + x^2/sqrt(x) = 4x^(-1/2) + x^(3/2)
Antiderivative: 4 · 2x^(1/2) + 2x^(5/2)/5 = 8sqrt(x) + 2x^(5/2)/5
Evaluate at 4: 8(2) + 2(32)/5 = 16 + 64/5 = 80/5 + 64/5 = 144/5
Evaluate at 1: 8(1) + 2(1)/5 = 8 + 2/5 = 42/5
Subtract: 144/5 - 42/5 = 102/5
Simplify the integrand: v^6/v^5 + 4v^8/v^5 = v + 4v^3
Antiderivative: v^2/2 + v^4
Evaluate at 2: 4/2 + 16 = 2 + 16 = 18
Evaluate at 1: 1/2 + 1 = 3/2
Subtract: 18 - 3/2 = 36/2 - 3/2 = 33/2
Power rule: x^(4/7 + 1) / (4/7 + 1) = x^(11/7) / (11/7) = 7x^(11/7) / 11
Evaluate at 1: 7(1)/11 = 7/11
Evaluate at 0: 0
Result: 7/11
Fractional exponents follow exactly the same power rule. Add 1 to the exponent (4/7 + 1 = 11/7), then divide by the new exponent.
Antiderivative of sin(theta) = -cos(theta)
Evaluate at pi: -cos(pi) = -(-1) = 1
Evaluate at pi/4: -cos(pi/4) = -(sqrt(2)/2) = -sqrt(2)/2
Subtract: 1 - (-sqrt(2)/2) = 1 + sqrt(2)/2 = (2 + sqrt(2))/2
Result: (2 + sqrt(2))/2
The negative sign in the antiderivative of sin is where most errors happen. Double-check: d/dx[-cos(x)] = sin(x). Correct.
Pull the constant: 5 times the integral of 1/(1 + x^2) dx
Antiderivative: 5 arctan(x)
Evaluate at sqrt(3): 5 arctan(sqrt(3)) = 5(pi/3)
Evaluate at 1/sqrt(3): 5 arctan(1/sqrt(3)) = 5(pi/6)
Subtract: 5(pi/3) - 5(pi/6) = 5(2pi/6 - pi/6) = 5(pi/6)
Result: 5pi/6
You must know these arctan values from memory: arctan(1/sqrt(3)) = pi/6, arctan(1) = pi/4, arctan(sqrt(3)) = pi/3. They come directly from the unit circle.
Students try to integrate (u + 3)(u - 4) without expanding first. You cannot apply the power rule to a product; expand, then integrate term by term.
Mishandling (9)^(3/2). This means take the square root first, then cube: sqrt(9) = 3, then 3^3 = 27. Students sometimes cube first and then try to take a square root of 729, which is correct but harder.
Forgetting to rewrite 1/sqrt(x) as x^(-1/2) before integrating. The power rule requires an explicit exponent.
Sign errors with negative lower bounds. In Problem 14, (-1)^4 = 1, not -1. Even exponents make negatives positive; odd exponents preserve the sign.
Confusing arctan values with sin/cos values. arctan(sqrt(3)) = pi/3 (not sqrt(3)/2, which is sin(pi/3)). These are different functions with different outputs.
⚠️ "Simplify before integrating" problems (like 11 and 12) appear on nearly every exam. The integration is trivial once the algebra is done, and the algebra is where marks are earned.
⚠️ Fractional exponent problems (like 6 and 15) test whether you truly understand the power rule or just memorise integer cases.
⚠️ The arctan integral (Problem 13) is a common exam item because it tests recognition: you have to spot the 1/(1 + x^2) form and recall the correct antiderivative.
⚠️ Negative results (Problem 8) are fine. If the function is below the x-axis, the integral is negative. Students sometimes re-do the problem thinking they made an error.
⚠️ Fraction arithmetic is where most marks are lost in definite-integral problems. Write every step when finding common denominators.
Q: Evaluate the integral from 1 to 4 of (3x^2 - 2x + 1) dx.
A: Antiderivative is x^3 - x^2 + x. At 4: 64 - 16 + 4 = 52. At 1: 1 - 1 + 1 = 1. Result: 52 - 1 = 51.
Q: Evaluate the integral from 1 to 9 of (1/sqrt(x)) dx.
A: Rewrite as x^(-1/2). Antiderivative: 2x^(1/2) = 2sqrt(x). At 9: 2(3) = 6. At 1: 2(1) = 2. Result: 6 - 2 = 4.
Q: Evaluate the integral from 0 to pi/2 of cos(theta) d(theta).
A: Antiderivative: sin(theta). At pi/2: sin(pi/2) = 1. At 0: sin(0) = 0. Result: 1.
Q: Simplify (8x^3 + 2x) / (2x), then integrate from 1 to 3.
A: Simplify to 4x^2 + 1. Antiderivative: 4x^3/3 + x. At 3: 36 + 3 = 39. At 1: 4/3 + 1 = 7/3. Result: 39 - 7/3 = 117/3 - 7/3 = 110/3.
Q: Evaluate the integral from 0 to 1 of 3/(1 + x^2) dx.
A: 3 arctan(x). At 1: 3(pi/4). At 0: 0. Result: 3pi/4.
Q: Evaluate the integral from -2 to 2 of x^5 dx.
A: Antiderivative: x^6/6. At 2: 64/6 = 32/3. At -2: (-2)^6/6 = 64/6 = 32/3. Result: 32/3 - 32/3 = 0. (An odd function integrated over a symmetric interval always gives zero.)
Q: Evaluate the integral from 0 to 1 of (t + 1)^2 dt.
A: Expand: t^2 + 2t + 1. Antiderivative: t^3/3 + t^2 + t. At 1: 1/3 + 1 + 1 = 7/3. At 0: 0. Result: 7/3.
Q: Evaluate the integral from 1 to 8 of x^(2/3) dx.
A: Power rule: x^(5/3) / (5/3) = 3x^(5/3)/5. At 8: 3(32)/5 = 96/5. At 1: 3/5. Result: 96/5 - 3/5 = 93/5.
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