Practice set
Applications of Integrals practice quiz 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 between \(y = 1 - x^2\) and the x-axis from \(x = -1\) to \(x = 1\)?
Correct answer: D. \(4/3\)
Explanation: Area = \(\int_{-1}^{1} (1 - x^2) dx = [x - x^3/3]_{-1}^{1} = (1 - 1/3) - (-1 + 1/3) = 4/3\).
What is the area under the curve \(y = 3x^2\) from \(x = 0\) to \(x = 2\)?
Correct answer: A. \(8\)
Explanation: Integral of \(3x^2\) is \(x^3\). From 0 to 2: \(2^3 = 8\).
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: \(V = 2\pi \int_0^2 x\cdot x \,dx = 2\pi \int_0^2 x^2 \,dx = \tfrac{16\pi}{3}\).
Using the disk method, find the volume of the solid obtained by rotating the region under \(y=\sqrt{x}\) from \(x=0\) to \(x=9\) about the x-axis.
Correct answer: C. \(81\pi/2\)
Explanation: Volume = \(\pi\int_0^9 (\sqrt{x})^2 dx = \pi\int_0^9 x dx = 81\pi/2\).
What is the area between the curves \(y = x^2\) and \(y = 1\) from \(x = -1\) to \(x = 1\)?
Correct answer: A. \(4/3\)
Explanation: Integrate \(1 - x^2\): \(\int_{-1}^1 (1 - x^2) \,dx = \bigl[x - x^3/3\bigr]_{-1}^1 = (1 - 1/3) - (-1 + 1/3) = 4/3\).
What is the area under the curve \(y = 3x\) from \(x = 1\) to \(x = 3\)?
Correct answer: A. \(12\)
Explanation: Integral of \(3x\) is \(3x^2/2\). From 1 to 3: \((27/2) - (3/2) = 12\).
What is the volume obtained by the shell method when rotating the region under \(y = x\) from \(x=0\) to \(x=3\) about the y‑axis?
Correct answer: D. \(18\pi\)
Explanation: By shells: \[ V = 2\pi\int_0^3 x\cdot y\,dx = 2\pi\int_0^3 x^2\,dx = 2\pi\Bigl[\tfrac{x^3}{3}\Bigr]_0^3 = 2\pi\cdot9 = 18\pi. \]
What is the volume of the solid obtained by rotating the region between \(y = 2\) and \(y = x\) from \(x = 0\) to \(x = 2\) around the \(y\)-axis?
Correct answer: C. \(8\pi/3\)
Explanation: Shell method: \(V = 2\pi \int_0^2 x(2 - x) \,dx = 2\pi \cdot 4/3 = 8\pi/3\).
What is the average value of \(f(x)=3x\) on \([0,4]\)?
Correct answer: C. \(6\)
Explanation: Average = \((1/4)\int_0^4 3x dx = (1/4)(3·8)=6\).
Find the arc length of \(y=2x\) from \(x=0\) to \(x=3\).
Correct answer: C. \(3\sqrt{5}\)
Explanation: Slope 2: length = \(\int_0^3 \sqrt{1+4} dx = 3\sqrt{5}\).
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Learn Applications of Integrals: Interactive Problems and Worked Examples
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

