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Quizzes > Mathematics > Advanced Math

Applications of the Definite Integral: Can You Ace the Quiz?

Ready for a Definite Integrals Quiz? Dive into Real-World Application Problems

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art of mathematical curves and solids on a yellow background for a quiz on definite integral area volume curve length

Ready to deepen your understanding of applications of the definite integral? In this free scored quiz, you'll put your skills to the test with area, volume, and curve length challenges designed for students passionate about applications of integrals. Whether you're sharpening up for a calculus class or hunting for a solid definite integrals quiz, this calculus application quiz will help you measure your strengths and uncover integral application problems you've yet to master. Curious about more questions about integration? Check out our questions about integration or explore higher-level advanced calculus problems . Jump in and start now!

What is the area under the curve y = x² from x = 0 to x = 2?
8/3
4
2
2/3
The area is given by ??² x² dx = [x³/3]?² = 8/3. This uses the power rule for integration. See a detailed example at Khan Academy.
Find the volume of the solid obtained by rotating the region under y = ?x from x = 0 to x = 4 about the x-axis.
4?
8?
16?
2?
Volume by disks: V = ???? (?x)² dx = ???? x dx = ?[x²/2]?? = 8?. This is a standard disk method application. More details at Lamar University Tutorial.
What is the length of the line segment of y = 3x from x = 0 to x = 1?
?2
?5
?10
3
Length L = ??¹ ?(1 + (dy/dx)²) dx = ??¹ ?(1 + 9) dx = ??¹ ?10 dx = ?10. See the formula at MathWorld.
Compute the average value of f(x) = x on the interval [0, 2].
1
4
1/2
2
Average value = (1/(2?0)) ??² x dx = (1/2)[x²/2]?² = (1/2)(2) = 1. See more at MIT OpenCourseWare.
Find the area between the horizontal line y = 2 and the x-axis from x = 1 to x = 3.
2
4
8
6
Area = ??³ 2 dx = 2(x)??³ = 4. This is a rectangle of height 2 and width 2. For reference, see Math Is Fun.
What is the area of the region bounded by y = x² and y = x from x = 0 to x = 1?
1/4
1/3
1/6
1/2
Area = ??¹ (x ? x²) dx = [x²/2 ? x³/3]?¹ = 1/2 ? 1/3 = 1/6. See similar examples at Khan Academy.
Using the cylindrical shell method, find the volume of the solid generated by rotating y = x from x = 0 to x = 2 about the y-axis.
4?/3
2?
6?
8?/3
Shell radius = x, height = x, so V = 2???² x·x dx = 2???² x² dx = 16?/3. See method at Lamar University.
A 10 m chain of linear mass density 2 kg/m is lifted vertically from the ground. What is the work done? (Use g = 9.8 m/s².)
1470 J
490 J
1960 J
980 J
Work = ??¹? (2·9.8·x) dx = 2·9.8·(x²/2)??¹? = 980 J. This uses W = ? weight·distance. See details at The Physics Classroom.
Find the average value of f(x) = sin x over [0, ?].
2
1/?
2/?
1
Average value = (1/?)??^? sin x dx = (1/?)[?cos x]?^? = 2/?. See explanation at Lamar University.
Compute the length of the curve y = ln x from x = 1 to x = e.
??? ?(1 + 1/x²) dx
?2
e ? 1
?2 + ln(e + 1)
Exact length is L = ??^e ?(1 + (1/x)²) dx; it cannot be expressed by elementary functions. The integral form is the correct representation. More at MathWorld.
Find the surface area of the solid formed by rotating y = x³ from x = 0 to x = 2 about the x-axis.
2???² x³?(1 + 9x?) dx
2???² x?(1 + 9x?) dx
4???² x³ dx
???² (x³)² dx
Surface area S = 2???² y?(1 + (dy/dx)²) dx = 2???² x³?(1 + 9x?) dx. This applies the surface area formula. See steps at Lamar University.
Determine the length of the parametric curve given by x = t², y = t³ for 0 ? t ? 1.
??¹ (2t + 3t²) dt
??¹ ?(1 + (3t²)²) dt
?13
??¹ ?(4t² + 9t?) dt
Length L = ??¹ ?((dx/dt)² + (dy/dt)²) dt = ??¹ ?(4t² + 9t?) dt. Parametric arc length uses both derivatives. More at MathWorld.
A conical tank with height 10 m and radius 5 m is filled with water (density 1000 kg/m³). Compute the work required to pump all the water to the top.
W = 2?·1000·9.8 ??¹? (r(y))² (10 ? y) dy
W = ?·1000·9.8 ??¹? (r(y))² dy
W = 1000·9.8·5·10
W = 2? ??? y dy
Using slicing: r(y) = (5/10)y gives slice area ?[r(y)]², weight = density·g·?[r(y)]² dy, distance = 10?y. Combined: W = 2?·1000·9.8 ??¹? (y²/4)(10 ? y) dy. See method at Lamar University.
Find the area enclosed by one loop of the polar curve r = 2 + 3 cos ?.
??^{2?} (2 + 3 sin ?)² d?
½ ?_{??/2}^{?/2} (2 + 3 cos ?)² d?
½ ??^{?} (2 + 3 cos ?) d?
2?(2 + 3)
Area = ½? (r(?))² d? over the loop. For r = 2 + 3 cos ?, one loop is from ??/2 to ?/2, so area = ½?_{??/2}^{?/2}(2 + 3 cos ?)² d?. See details at MathWorld.
What is the surface area of a torus generated by rotating a circle of radius a whose center is at distance R (R > a) from the axis of rotation?
4?a²
4?²aR
2?²a²R
2?aR²
The surface area of a torus is S = (2?R)(2?a) = 4?²aR by Pappus's theorem or by integrating 2?y ds. Reference: MathWorld.
0
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Study Outcomes

  1. Analyze Area Between Curves -

    Use definite integrals to calculate and interpret the area enclosed by two or more functions, reinforcing your understanding of applications of the definite integral.

  2. Compute Solid Volumes -

    Apply methods like the disk, washer, and shell techniques to determine volumes of revolution, mastering integral application problems in three dimensions.

  3. Determine Curve Length -

    Calculate the arc length of given functions using definite integrals, honing your skills in integral calculus applications to measure curved paths.

  4. Apply Integrals to Practical Scenarios -

    Translate real-world situations into integral expressions and solve them, demonstrating how applications of integrals model physical and engineering problems.

  5. Assess Problem-Solving Proficiency -

    Engage with a scored definite integrals quiz to identify strengths and target areas for improvement in calculus application quiz challenges.

  6. Strengthen Calculus Application Skills -

    Reinforce core techniques through varied integral application problems, preparing you for exams and real-life uses of the definite integral.

Cheat Sheet

  1. Area Between Curves -

    Compute the area trapped between two functions by integrating the difference: A = ∫₝ᵇ [f(x) − g(x)] dx, where f(x) ≥ g(x). For example, the area between y = x and y = x² from 0 to 1 is ∫₀¹ (x − x²) dx = 1/6. Remember the "top minus bottom" rule popularized in MIT OpenCourseWare to avoid sign errors.

  2. Volume by Disks and Washers -

    Use the disk method when revolving a region around an axis without a gap: V = π ∫₝ᵇ [R(x)]² dx. If there's a hole, switch to washers: V = π ∫₝ᵇ ([R(x)]² − [r(x)]²) dx, as shown in Stewart's Calculus examples. Visualizing cross-sections as coins or washers can help you recall which radii go where.

  3. Volume by Cylindrical Shells -

    The shell method uses V = 2π ∫₝ᵇ x·f(x) dx when revolving vertical strips around the y-axis (swap roles for horizontal strips). For instance, revolving y = √x from x = 0 to 4 around the y-axis yields V = 2π ∫₀❴ x·√x dx = 2π·8 = 16π. A handy mnemonic is "shells shell out 2π·radius·height·thickness."

  4. Arc Length of a Curve -

    Calculate the length of y = f(x) over [a,b] via L = ∫₝ᵇ √[1 + (f′(x))²] dx, as detailed in Paul's Online Math Notes. For example, the length of y = x³/3 from 0 to 1 is ∫₀¹ √[1 + x❴] dx, which may require substitution or numerical methods. Recall "L is Long" to link length L with the square-root integrand.

  5. Surface Area of Revolution -

    Find the area of a surface formed by revolving y = f(x) about the x-axis using SA = 2π ∫₝ᵇ f(x) √[1 + (f′(x))²] dx. For y = sin(x) from 0 to π, SA = 2π ∫₀^π sin(x)√[1 + cos²(x)] dx, which appears in engineering applications. Use the "surface sizzle" trick: 2π·radius·arc-length element for memorable setup.

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