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10 FE practice problems: calculus: derivatives and integrals, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice calculus: derivatives and integrals, a topic from the FE exam specification, using the Mathematics, Integral Calculus part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.

How to use these problems

Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.

The problems

Problem 1 · FE, Calculus: derivatives and integrals

The area under the curve y = 3x² - 3x + 19 between x = 1 and x = 5 (the curve is above the x-axis there) is most nearly:

  • A 196
  • B 24.0
  • C 164
  • D 183

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
C (164)
Given
a = 3, b = -3, c = 19, x1 = 1, x2 = 5
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(3x² - 3x + 19) dx = (3/3)x³ + (-3/2)x² + 19x
  2. F(5) = 182.5; F(1) = 18.50
  3. Area = F(5) - F(1) = 164
Result
164, 3 significant figures
Check
the curve stays between y = 19.0 and y = 79.0 on the interval, so the area must lie between 76.0 and 316.
Why the others are wrong
  • A: used one trapezoid (average of the end heights times the width), not the integral
  • B: differentiated instead of integrating
  • D: evaluated the antiderivative at the upper limit only

Problem 2 · FE, Calculus: derivatives and integrals

The area under the curve y = 2x² + 16 between x = 3 and x = 7 (the curve is above the x-axis there) is most nearly:

  • A 296
  • B 341
  • C 696
  • D 275

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
D (275)
Given
a = 2, b = 0, c = 16, x1 = 3, x2 = 7
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(2x² + 16) dx = (2/3)x³ + (0/2)x² + 16x
  2. F(7) = 340.7; F(3) = 66.00
  3. Area = F(7) - F(3) = 275
Result
275, 3 significant figures
Check
the curve stays between y = 34.0 and y = 114 on the interval, so the area must lie between 136 and 456.
Why the others are wrong
  • A: used one trapezoid (average of the end heights times the width), not the integral
  • B: evaluated the antiderivative at the upper limit only
  • C: raised each power but did not divide by the new power

Problem 3 · FE, Calculus: derivatives and integrals

For f(x) = 3x³ + 8x² + 5x - 4, the slope of the curve at x = 3 is most nearly:

  • A 56.0
  • B 129
  • C 134
  • D 70.0

Handbook: Mathematics, Differential Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
C (134)
Given
a = 3, b = 8, c = 5, d = -4, x0 = 3
Find
f′(3)
Handbook
Mathematics, Differential Calculus (the derivative), page 47
Equation
d(axⁿ)/dx = n·a·xⁿ⁻¹; the slope is f′(x0)
Substitute
  1. f′(x) = 9x² + 16x + 5
  2. f′(3) = 3(3)(3)² + 2(8)(3) + (5) = 134
Result
134, 3 significant figures
Check
the slope of the curve is the first derivative evaluated at the point; the constant term drops out.
Why the others are wrong
  • A: lowered each power but did not multiply by it
  • B: dropped the derivative of the linear term
  • D: is the second derivative f″(x0)

Problem 4 · FE, Calculus: derivatives and integrals

The area under the curve y = 2x² + x + 13 between x = 1 and x = 6 (the curve is above the x-axis there) is most nearly:

  • A 530
  • B 20.0
  • C 268
  • D 226

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
D (226)
Given
a = 2, b = 1, c = 13, x1 = 1, x2 = 6
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(2x² + x + 13) dx = (2/3)x³ + (1/2)x² + 13x
  2. F(6) = 240.0; F(1) = 14.17
  3. Area = F(6) - F(1) = 226
Result
226, 3 significant figures
Check
the curve stays between y = 16.0 and y = 91.0 on the interval, so the area must lie between 80.0 and 455.
Why the others are wrong
  • A: raised each power but did not divide by the new power
  • B: differentiated instead of integrating
  • C: used one trapezoid (average of the end heights times the width), not the integral

Problem 5 · FE, Calculus: derivatives and integrals

The area under the curve y = 4x² - 5x + 20 between x = 0 and x = 3 (the curve is above the x-axis there) is most nearly:

  • A 123
  • B 91.5
  • C 73.5
  • D 24.0

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
C (73.5)
Given
a = 4, b = -5, c = 20, x1 = 0, x2 = 3
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(4x² - 5x + 20) dx = (4/3)x³ + (-5/2)x² + 20x
  2. F(3) = 73.50; F(0) = 0
  3. Area = F(3) - F(0) = 73.5
Result
73.5, 3 significant figures
Check
the curve stays between y = 18.4 and y = 41.0 on the interval, so the area must lie between 55.3 and 123.
Why the others are wrong
  • A: raised each power but did not divide by the new power
  • B: used one trapezoid (average of the end heights times the width), not the integral
  • D: differentiated instead of integrating

Problem 6 · FE, Calculus: derivatives and integrals

For f(x) = -x³ - 9x² - 4x + 2, the slope of the curve at x = -4 is most nearly:

  • A 16.0
  • B 20.0
  • C 6.00
  • D 24.0

Handbook: Mathematics, Differential Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
B (20.0)
Given
a = -1, b = -9, c = -4, d = 2, x0 = -4
Find
f′(-4)
Handbook
Mathematics, Differential Calculus (the derivative), page 47
Equation
d(axⁿ)/dx = n·a·xⁿ⁻¹; the slope is f′(x0)
Substitute
  1. f′(x) = -3x² - 18x - 4
  2. f′(-4) = 3(-1)(-4)² + 2(-9)(-4) + (-4) = 20
Result
20.0, 3 significant figures
Check
the slope of the curve is the first derivative evaluated at the point; the constant term drops out.
Why the others are wrong
  • A: lowered each power but did not multiply by it
  • C: is the second derivative f″(x0)
  • D: dropped the derivative of the linear term

Problem 7 · FE, Calculus: derivatives and integrals

For f(x) = 3x³ - 2x² - 6x - 8, the slope of the curve at x = 2 is most nearly:

  • A 44.0
  • B 2.00
  • C 28.0
  • D 22.0

Handbook: Mathematics, Differential Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
D (22.0)
Given
a = 3, b = -2, c = -6, d = -8, x0 = 2
Find
f′(2)
Handbook
Mathematics, Differential Calculus (the derivative), page 47
Equation
d(axⁿ)/dx = n·a·xⁿ⁻¹; the slope is f′(x0)
Substitute
  1. f′(x) = 9x² - 4x - 6
  2. f′(2) = 3(3)(2)² + 2(-2)(2) + (-6) = 22
Result
22.0, 3 significant figures
Check
the slope of the curve is the first derivative evaluated at the point; the constant term drops out.
Why the others are wrong
  • A: multiplied by each power but did not lower it
  • B: lowered each power but did not multiply by it
  • C: dropped the derivative of the linear term

Problem 8 · FE, Calculus: derivatives and integrals

The area under the curve y = 4x² + 2x + 6 between x = 2 and x = 5 (the curve is above the x-axis there) is most nearly:

  • A 528
  • B 213
  • C 195
  • D 348

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
C (195)
Given
a = 4, b = 2, c = 6, x1 = 2, x2 = 5
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(4x² + 2x + 6) dx = (4/3)x³ + (2/2)x² + 6x
  2. F(5) = 221.7; F(2) = 26.67
  3. Area = F(5) - F(2) = 195
Result
195, 3 significant figures
Check
the curve stays between y = 26.0 and y = 116 on the interval, so the area must lie between 78.0 and 348.
Why the others are wrong
  • A: raised each power but did not divide by the new power
  • B: used one trapezoid (average of the end heights times the width), not the integral
  • D: multiplied the height at the upper limit by the width

Problem 9 · FE, Calculus: derivatives and integrals

For f(x) = x³ + 9x² - 8x + 6, the slope of the curve at x = 1 is most nearly:

  • A 8.00
  • B 13.0
  • C 24.0
  • D 21.0

Handbook: Mathematics, Differential Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
B (13.0)
Given
a = 1, b = 9, c = -8, d = 6, x0 = 1
Find
f′(1)
Handbook
Mathematics, Differential Calculus (the derivative), page 47
Equation
d(axⁿ)/dx = n·a·xⁿ⁻¹; the slope is f′(x0)
Substitute
  1. f′(x) = 3x² + 18x - 8
  2. f′(1) = 3(1)(1)² + 2(9)(1) + (-8) = 13
Result
13.0, 3 significant figures
Check
the slope of the curve is the first derivative evaluated at the point; the constant term drops out.
Why the others are wrong
  • A: is f(x0), the function value, not the slope
  • C: is the second derivative f″(x0)
  • D: dropped the derivative of the linear term

Problem 10 · FE, Calculus: derivatives and integrals

The area under the curve y = 2x² - x + 11 between x = 0 and x = 4 (the curve is above the x-axis there) is most nearly:

  • A 100
  • B 156
  • C 78.7
  • D 16.0

Handbook: Mathematics, Integral Calculus, FE Reference Handbook 10.6

Show the worked solution
Answer
C (78.7)
Given
a = 2, b = -1, c = 11, x1 = 0, x2 = 4
Find
area under the curve
Handbook
Mathematics, Integral Calculus (definite integral), page 49
Equation
Area = ∫ f(x) dx from x1 to x2 = F(x2) - F(x1)
Substitute
  1. F(x) = ∫(2x² - x + 11) dx = (2/3)x³ + (-1/2)x² + 11x
  2. F(4) = 78.67; F(0) = 0
  3. Area = F(4) - F(0) = 78.7
Result
78.7, 3 significant figures
Check
the curve stays between y = 10.9 and y = 39.0 on the interval, so the area must lie between 43.5 and 156.
Why the others are wrong
  • A: used one trapezoid (average of the end heights times the width), not the integral
  • B: multiplied the height at the upper limit by the width
  • D: differentiated instead of integrating

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is calculus: derivatives and integrals in the FE Reference Handbook?

Look in the Mathematics, Integral Calculus part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.

Are these real FE exam questions?

No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.

How do I check my answer before opening the solution?

Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.

Sources

  1. NCEES FE Civil CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  3. NCEES FE Mechanical CBT exam specifications (PDF). Retrieved October 3, 2026.
  4. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.