Area
With vertical slices, subtract bottom from top.
Calculus Practice
Practice area, volume, average value, accumulated change, displacement, total distance, and other applications of definite integrals.
Try each problem first, then click Show solution to check your setup and reasoning.
Keep These Setups Nearby
With vertical slices, subtract bottom from top.
Outer disk minus inner disk.
Radius times height times circumference factor \(2\pi\).
Level 1
Practice area under curves, signed area, and total geometric area.
Problem 1
Area Under a CurveStep-by-Step Solution
Because the function is above the \(x\)-axis,
\[ A = \int_0^2x^2\,dx. \]Problem 2
Signed AreaStep-by-Step Solution
\[ \int_0^4f(x)\,dx = \int_0^2f(x)\,dx + \int_2^4f(x)\,dx. \]Problem 3
Total AreaStep-by-Step Solution
Total geometric area counts both regions positively.
Level 2
Identify bounds, determine which function is on top, and build the correct area integral.
Problem 4
IntersectionsStep-by-Step Solution
Set the equations equal:
\[ 2x=x^2. \] \[ x(x-2)=0. \]Problem 5
Area Between CurvesStep-by-Step Solution
The curves intersect at \(x=0\) and \(x=2\).
On this interval, \(2x\) is above \(x^2\).
\[ A = \int_0^2 (2x-x^2)\,dx. \]Problem 6
Horizontal SlicesStep-by-Step Solution
A horizontal slice extends from the left boundary to the right boundary.
Level 3
Practice disks, washers, and cylindrical shells.
Problem 7
Disk MethodStep-by-Step Solution
The radius is
\[ R(x)=x. \]Use the disk formula:
\[ V = \pi \int_0^2x^2\,dx. \]Problem 8
Washer MethodStep-by-Step Solution
\[ V = \pi \int_0^2 \left( 3^2-x^2 \right) dx. \]Problem 9
Shell MethodStep-by-Step Solution
Radius:
\[ r=x. \]Height:
\[ h=x. \] \[ V = 2\pi \int_0^2x(x)\,dx. \]Level 4
Use integrals to calculate average values and accumulated quantities.
Problem 10
Average ValueStep-by-Step Solution
\[ f_{\text{avg}} = \frac{1}{2-0} \int_0^2x^2\,dx. \]Problem 11
Net ChangeStep-by-Step Solution
Integrate the rate over the time interval.
\[ \int_0^3(4t+2)\,dt. \]Problem 12
Accumulation FunctionStep-by-Step Solution
By the Fundamental Theorem of Calculus,
\[ \frac{d}{dx} \left[ \int_a^xf(t)\,dt \right] = f(x). \]Level 5
Distinguish displacement from distance and choose the correct integral setup.
Problem 13
DisplacementStep-by-Step Solution
\[ \text{Displacement} = \int_0^4(2t-4)\,dt. \]Problem 14
Total DistanceStep-by-Step Solution
Velocity changes sign when
\[ 2t-4=0 \Rightarrow t=2. \]Split the interval:
\[ \text{Distance} = - \int_0^2(2t-4)\,dt + \int_2^4(2t-4)\,dt. \]Problem 15
Method SelectionStep-by-Step Solution
Vertical slices are parallel to the \(y\)-axis. Rotating them around the \(y\)-axis creates cylindrical shells.
Problem 16
Mixed ReviewStep-by-Step Solution
Divide the accumulated value by the length of the interval.
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