Applications of Integrals Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below to practice applications of integrals — the most important "real use" skills from Calculus: area under a curve and definite integrals \(\int_a^b f(x)\,dx\) as net or total accumulation, area between curves using top minus bottom (or right minus left), volume of solids of revolution with the disk method and washer method about the \(x\)-axis or \(y\)-axis, the shell method for rotation when washers are awkward, and surface area of revolution using arc length factors like \(\sqrt{1+(f'(x))^2}\). You will learn how to sketch the region, find intersection points, choose correct bounds, and write the correct integral setup with the right radii, heights, and units. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
How this applications of integrals practice works
- 1. Take the practice set: answer the area, volume, and surface area questions below.
- 2. Open the lesson (optional): review area under the curve, area between curves, disk/washer/shell volume methods, and surface area of revolution.
- 3. Retry: return to the question set and set up the correct integral formula immediately.
What you will learn in the applications of integrals lesson
Area with definite integrals
- Area under a curve when \(f(x)\ge 0\): \(\displaystyle A=\int_a^b f(x)\,dx\)
- Total area vs net area when a function crosses the axis
- Area between curves: \(\displaystyle A=\int_a^b(\text{top}-\text{bottom})\,dx\)
Volume: disk and washer methods
- Disk method (solid region): \(\displaystyle V=\pi\int_a^b [R(x)]^2\,dx\)
- Washer method (hole inside): \(\displaystyle V=\pi\int_a^b\big([R(x)]^2-[r(x)]^2\big)\,dx\)
- Rotate about the \(x\)-axis or \(y\)-axis with correct radii and bounds
Volume: cylindrical shells
- Shell method: \(\displaystyle V=2\pi\int (\text{radius})(\text{height})\,dx\) or \(dy\)
- Use shells when rotating about the \(y\)-axis with \(x\)-slices (or when washers require solving for \(x\) in terms of \(y\))
- Set radius as distance to the axis of rotation, height as curve difference
Surface area and setup skills
- Surface area of revolution: \(\displaystyle S=2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}\,dx\)
- Find intersection points by solving equations like \(2x=x^2\)
- Always verify units: area in square units, volume in cubic units
Practice set
Applications of Integrals practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
What is the area under the curve \(y=1\) from \(x=0\) to \(x=2\)?
Correct answer: A. \(2\)
Explanation: Integral of 1 is \(x\). Evaluated from 0 to 2 gives \(2 - 0 = 2\).
What is the volume of the solid obtained by rotating the line segment \(y=x\) from \(x=0\) to \(x=1\) around the \(x\)-axis?
Correct answer: B. \(\pi/3\)
Explanation: Disk method: \(V = \pi \int_0^1 x^2 \,dx = \pi\bigl[ x^3/3\bigr]_0^1 = \pi/3\).
What is the area under the curve \(y=x\) from \(x=0\) to \(x=3\)?
Correct answer: B. \(9/2\)
Explanation: Integral of \(x\) is \(x^2/2\). From 0 to 3: \(9/2 - 0 = 9/2\).
What is the area under the curve \(y=2\) from \(x=0\) to \(x=4\)?
Correct answer: B. \(8\)
Explanation: Integral of constant 2 is \(2x\). Evaluated from 0 to 4 gives \(8\).
What is the area under the curve \(y=x\) from \(x=0\) to \(x=5\)?
Correct answer: B. \(25/2\)
Explanation: Integral of \(x\) is \(x^2/2\). From 0 to 5: \(25/2\).
What is the area between the curves \(y=1\) and \(y=x\) from \(x=0\) to \(x=1\)?
Correct answer: A. \(1/2\)
Explanation: Integrate \(1 - x\): \([x - x^2/2]_0^1 = 1/2\).
What is the volume of the solid obtained by rotating \(y=\sqrt{x}\) from \(x=0\) to \(x=1\) around the \(x\)-axis?
Correct answer: A. \(\pi/2\)
Explanation: Disk method: \(V = \pi \int_0^1 (\sqrt{x})^2 \,dx = \pi \int_0^1 x \,dx = \pi/2\).
What is the volume of the solid obtained by rotating the area under \(y = 1\) from \(x = 0\) to \(x = 2\) around the \(x\)-axis?
Correct answer: C. \(2\pi\)
Explanation: Disk method: \(V = \pi \int_0^2 1^2 \,dx = 2\pi\).
What is the volume of the solid obtained by rotating the region between \(y = 2\) and \(y = 1\) from \(x = 0\) to \(x = 1\) around the \(x\)-axis?
Correct answer: A. \(3\pi\)
Explanation: Washer method: \(V = \pi \int_0^1 (2^2 - 1^2) \,dx = 3\pi\).
What is the volume of the solid obtained by rotating the region under \(y = x\) from \(x = 0\) to \(x = 2\) about the \(y\)-axis?
Correct answer: D. \(16\pi/3\)
Explanation: Shell method: \(V = 2\pi \int_0^2 x\cdot x \,dx = 2\pi \int_0^2 x^2 \,dx = \tfrac{16\pi}{3}\).
Result
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