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8.1 Finding the Average Value of a Function on an Interval

For a continuous function on $[a,b]$ , the average value is its total signed accumulation divided by the interval length. It is the constant height whose rectangle of width $b-a$ has the same signed area as the integral of $f$ .

f_{\mathrm{avg}}=\frac{1}{b-a}\int_a^b f(x),dx

Identify the complete interval $[a,b]$ .
Evaluate $\int_a^b f(x)\,dx$ .
Divide by the positive interval length $b-a$ .
Report the result in the same units as $f$ .

For $f(x)=x^2$ on $[0,3]$ , $f_{\mathrm{avg}}=\frac{1}{3-0}\int_0^3x^2\,dx=\frac13[x^3/3]_0^3=\frac13(9)=3$ . The accumulated value is $9$ function-units times input-units; dividing by the interval length $3$ leaves the units of $f$ .

Average value uses $\frac{1}{b-a}\int_a^b f(x)\,dx$ ; average rate of change uses $\frac{f(b)-f(a)}{b-a}$ . They answer different questions. Because the integral is signed, an average value may be zero or negative even when the graph encloses positive geometric area elsewhere.

8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

For motion along a line, $v(t)=s'(t)$ and $a(t)=v'(t)$ . Definite integrals reverse these derivative relationships: accumulated acceleration changes velocity, and accumulated velocity changes position.

v(b)=v(a)+\int_a^b a(t),dt,\qquad s(b)=s(a)+\int_a^b v(t),dt

Quantity on $[a,b]$	Calculation	Sign meaning
displacement	$\int_a^b v(t)\,dt$	forward and backward motion cancel
final position	$s(a)+\int_a^b v(t)\,dt$	initial position plus displacement
total distance	$\int_a^b\|v(t)\|\,dt$	all traveled lengths are positive

Let $v(t)=t-2$ meters/second for $0\le t\le4$ and $s(0)=5$ meters. Displacement is $\int_0^4(t-2)\,dt=0$ , so $s(4)=5$ meters. Since velocity changes sign at $t=2$ , total distance is $\int_0^2(2-t)\,dt+\int_2^4(t-2)\,dt=2+2=4$ meters.

Do not use $|\int v|$ for total distance: cancellation has already occurred inside that integral. Split at every time when velocity changes sign, or integrate $|v|$ . If velocity is in meters/second and time in seconds, its definite integral is in meters—not meters/second.

8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

If $r(t)$ is the rate of change of a quantity, then $A(x)=\int_a^x r(t)\,dt$ records the quantity’s net change from input $a$ to input $x$ . The lower bound fixes the starting point; the upper bound makes the accumulated value change with $x$ .

A(x)=\int_a^x r(t),dt,\qquad A(a)=0,\qquad A'(x)=r(x)

Feature	Interpretation
$r(t)>0$	positive contributions make $A$ increase
$r(t)<0$	negative contributions make $A$ decrease
$\int_a^b r(t)\,dt$	net change from $a$ to $b$
rate units $Q$ /time	integral units $Q$

If the actual quantity is $Q$ and $Q(a)=Q_0$ , then $Q(x)=Q_0+A(x)=Q_0+\int_a^x r(t)\,dt$ . Thus the integral alone is the change since the start; adding the initial amount gives the current quantity.

Net change is signed. Positive and negative rate contributions can cancel, so it is not automatically the total amount of activity. Also, $A(a)=0$ does not mean the original quantity was zero; it means no change has accumulated over an interval of zero length.

In an applied accumulation problem, first identify the rate that changes the target quantity. When material enters and leaves, use net rate = incoming rate $-$ outgoing rate. Integrating that net rate gives the signed change, not the final amount by itself.

Q(b)=Q(a)+\int_a^b Q'(t),dt

Define the target quantity and its units.
Build the signed net rate in quantity/time units.
Integrate over the exact time interval.
Interpret the integral as net change.
Add the initial quantity if the question asks for the amount at the end.

A tank initially contains $100$ liters and has illustrative net inflow $r(t)=12-2t$ liters/minute for $0\le t\le4$ . Its net change is $\int_0^4(12-2t)\,dt=[12t-t^2]_0^4=32$ liters. Therefore the amount after $4$ minutes is $100+32=132$ liters.

Do not add the initial amount when only net change is requested, and do not omit it when the final amount is requested. Check that rate and integration-variable units match; integrating liters/minute with respect to minutes produces liters.

8.4 Finding the Area Between Curves Expressed as Functions of x

When curves are written as functions of $x$ , use a vertical slice of width $dx$ . If $y=u(x)$ is above $y=\ell(x)$ from $x=a$ to $x=b$ , the slice height is $u(x)-\ell(x)$ , so integrating those rectangle areas gives the region’s area.

A=\int_a^b\big(\text{upper}(x)-\text{lower}(x)\big),dx

Find the left and right boundaries, often by solving the curve-intersection equation.
Test the interval to identify the upper and lower functions.
Write upper minus lower.
Split the integral at any point where their order changes.
Evaluate and report square units.

The curves $y=2x$ and $y=x^2$ intersect where $x^2=2x$ , so $x=0$ and $x=2$ . On $(0,2)$ , $2x>x^2$ . Therefore $A=\int_0^2(2x-x^2)\,dx=[x^2-x^3/3]_0^2=4-8/3=4/3$ square units.

A negative integral signals that the curves were subtracted in the wrong order on part or all of the interval; geometric area is never negative. Do not use one upper-minus-lower expression across an intersection where the curve order switches—split there and write a nonnegative slice height on each piece.

8.5 Finding the Area Between Curves Expressed as Functions of y

When boundaries are written as $x$ -functions of $y$ , use a horizontal slice of thickness $dy$ . If $x=r(y)$ is the right boundary and $x=\ell(y)$ is the left boundary from $y=c$ to $y=d$ , the slice width is $r(y)-\ell(y)$ .

A=\int_c^d\big(\text{right}(y)-\text{left}(y)\big),dy

Integration variable	Slice	Difference	Bounds
$dx$	vertical	upper $-$ lower	$x$ -values
$dy$	horizontal	right $-$ left	$y$ -values

For the region bounded by $x=y^2$ and $x=2y$ , intersections satisfy $y^2=2y$ , so $y=0$ and $y=2$ . On $0<y<2$ , $2y$ is to the right of $y^2$ . Thus $A=\int_0^2(2y-y^2)\,dy=[y^2-y^3/3]_0^2=4/3$ square units.

A $dy$ integral must use expressions in $y$ and $y$ -value limits; do not mix them with $x$ -bounds. If a horizontal line meets different right or left curves in different vertical ranges, split the integral at the $y$ -value where that boundary changes.

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

If two curves intersect several times, their upper–lower order may change at each intersection. Use every relevant intersection as a possible partition point, test the sign of $f-g$ on each subinterval, and make each slice height nonnegative.

A=\sum_i\int_{x_i}^{x_{i+1}}(\text{upper}-\text{lower}),dx=\int_a^b|f(x)-g(x)|,dx

Solve $f(x)=g(x)$ for all intersections in the interval.
Order those $x$ -values.
Test one point in each subinterval.
Write upper minus lower on each piece, or retain one absolute-difference integral.
Add all nonnegative contributions.

On $[0,2\pi]$ , $y=\sin x$ and $y=0$ intersect at $0$ , $\pi$ , and $2\pi$ . Since $\sin x\ge0$ on $[0,\pi]$ and $\sin x\le0$ on $[\pi,2\pi]$ , $A=\int_0^\pi\sin x\,dx+\int_\pi^{2\pi}(-\sin x)\,dx=2+2=4$ . Equivalently, $A=\int_0^{2\pi}|\sin x|\,dx=4$ .

The unsplit signed integral $\int_0^{2\pi}\sin x\,dx=0$ is net signed area, not total geometric area. Absolute value must be applied to the height before integration, not to the final signed integral after positive and negative regions have already canceled.

8.7 Volumes with Cross Sections: Squares and Rectangles

For a solid whose cross sections perpendicular to the $x$ -axis are known, a thin slice of thickness $dx$ has volume approximately $A(x)\,dx$ . Adding all slices gives the exact volume. The base region supplies a length such as $s(x)=\text{top}-\text{bottom}$ ; the named cross-sectional shape determines how that length becomes area.

V=\int_a^b A(x),dx,\qquad A_{\text{square}}(x)=[s(x)]^2,\qquad A_{\text{rectangle}}(x)=\ell(x)w(x)

Identify the slicing direction and interval.
Express the required side length or lengths from the base region.
Use the square or rectangle area formula to create $A(x)$ .
Integrate $A(x)$ across the interval.
Report a nonnegative result in cubic units.

Example: the base is between $y=x$ and $y=x^2$ on $0\le x\le1$ , and perpendicular cross sections are squares. Here $s(x)=x-x^2$ , so $A(x)=(x-x^2)^2$ . Therefore $V=\int_0^1(x^2-2x^3+x^4)\,dx=\left[\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}\right]_0^1=\frac{1}{30}$ cubic units.

Do not integrate the slice length itself. A distance such as $s(x)$ has linear units; the integrand must be the cross-sectional area $A(x)$ , which has square units. Multiplying by slice thickness through integration then produces cubic units.

8.8 Volumes with Cross Sections: Triangles and Semicircles

Let $s(x)$ be the segment cut from the base region by a slice perpendicular to the $x$ -axis. The problem states what that segment represents—such as a triangle base, a triangle leg, or a semicircle diameter. First convert $s(x)$ into cross-sectional area $A(x)$ ; then accumulate those areas with $V=\int_a^b A(x)\,dx$ .

Cross section and meaning of $s$	Area function
Triangle with base $s$ and height $h(x)$	$A(x)=\frac12s(x)h(x)$
Equilateral triangle with side $s$	$A(x)=\frac{\sqrt3}{4}[s(x)]^2$
Semicircle with diameter $s$	$A(x)=\frac{\pi}{8}[s(x)]^2$

Find $s(x)$ from the base boundaries.
Read exactly what geometric dimension $s(x)$ represents.
Substitute it into the stated shape's area formula.
Integrate the resulting $A(x)$ over the base interval.
Check that the volume is nonnegative and has cubic units.

Example: the base lies between $y=x$ and $y=0$ for $0\le x\le2$ , and each perpendicular cross section is a semicircle whose diameter is the vertical segment. Thus $s(x)=x$ , so $A(x)=\frac{\pi}{8}x^2$ . Therefore $V=\int_0^2\frac{\pi}{8}x^2\,dx=\frac{\pi}{8}\left[\frac{x^3}{3}\right]_0^2=\frac{\pi}{3}$ cubic units.

For a semicircle, do not use the given diameter as the radius. If the base segment is the diameter $s$ , then $r=s/2$ and $A=\frac12\pi(s/2)^2=\pi s^2/8$ . Likewise, a triangle needs both base and height unless its type fixes their relationship.

8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

When a region touches the axis of rotation, a slice perpendicular to that axis sweeps out a solid circular disc. Its radius $R$ is the distance from the axis to the region's outer boundary, so a thin slice has volume approximately $\pi[R]^2$ times its thickness.

\text{about the }x\text{-axis: }V=\pi\int_a^b[R(x)]^2,dx,\qquad \text{about the }y\text{-axis: }V=\pi\int_c^d[R(y)]^2,dy

Draw slices perpendicular to the axis of rotation.
Use $dx$ for vertical discs about the $x$ -axis and $dy$ for horizontal discs about the $y$ -axis.
Express the radius as a nonnegative distance to the axis.
Integrate the disc area $\pi R^2$ over the stated bounds.

Example: rotate the region under $y=\sqrt{x}$ from $x=0$ to $x=4$ around the $x$ -axis. Each vertical slice forms a disc with $R(x)=\sqrt{x}$ . Therefore $V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4=8\pi$ cubic units.

Square the radius, not the original function automatically: the radius must first be identified as a distance to the chosen axis. The disc formula also assumes no central hole; if the region does not reach the axis of rotation, subtracting an inner circular area is required instead.

8.10 Volume with Disc Method: Revolving Around Other Axes

For rotation around a line other than a coordinate axis, the disc radius is not usually the function value. It is the perpendicular distance from the axis of rotation to the outer boundary. The region must reach the axis so each perpendicular slice forms a solid disc.

y=k:\ R(x)=|f(x)-k|,\quad V=\pi\int_a^b[R(x)]^2,dx;\qquad x=h:\ R(y)=|g(y)-h|,\quad V=\pi\int_c^d[R(y)]^2,dy

Mark the horizontal or vertical axis of rotation.
Choose slices perpendicular to it.
Write radius as outer boundary minus axis, or axis minus outer boundary, so it represents a nonnegative distance.
Square the radius, multiply by $\pi$ , and integrate over the matching variable's bounds.

Example: rotate the region between $y=-1$ and $y=x^2$ for $0\le x\le1$ about $y=-1$ . Each vertical slice forms a disc with $R(x)=x^2-(-1)=x^2+1$ . Thus $V=\pi\int_0^1(x^2+1)^2\,dx=\pi\left[\frac{x^5}{5}+\frac{2x^3}{3}+x\right]_0^1=\frac{28\pi}{15}$ cubic units.

Do not use $R=f(x)$ merely because the boundary is $y=f(x)$ ; that is valid only when the axis is $y=0$ . For a shifted axis, compute the distance first. If the rotated region leaves a central gap, the cross section is not a disc and this single-radius formula does not apply.

8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

If a region does not reach the axis of rotation, a perpendicular slice produces a washer rather than a solid disc. Let $R$ be the farther distance from the axis and $r$ the nearer distance, with $R\ge r\ge0$ . The washer area is the area of the outer circle minus the inner circular hole.

A=\pi(R^2-r^2);\qquad \text{about the }x\text{-axis: }V=\pi\int_a^b\big(R(x)^2-r(x)^2\big),dx;\qquad \text{about the }y\text{-axis: }V=\pi\int_c^d\big(R(y)^2-r(y)^2\big),dy

Use slices perpendicular to the rotation axis.\n2. Measure both boundary distances from that axis.\n3. Label the farther distance $R$ and the nearer distance $r$ .\n4. Integrate $\pi(R^2-r^2)$ over the appropriate bounds.\n5. Check that $R\ge r$ throughout each interval.

Example: rotate the region between $y=\sqrt{x}$ and $y=x^2$ on $0\le x\le1$ around the $x$ -axis. Since $\sqrt{x}\ge x^2$ , $R(x)=\sqrt{x}$ and $r(x)=x^2$ . Thus $V=\pi\int_0^1\left[(\sqrt{x})^2-(x^2)^2\right]dx=\pi\int_0^1(x-x^4)\,dx=\pi\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=\frac{3\pi}{10}$ cubic units.

A washer is a difference of two circular areas: $\pi R^2-\pi r^2=\pi(R^2-r^2)$ . It is not $\pi(R-r)^2$ . If the farther and nearer boundaries exchange, split the integral where their roles change so the cross-sectional area stays nonnegative.

8.12 Volume with Washer Method: Revolving Around Other Axes

A region rotated around a horizontal line $y=k$ or vertical line $x=h$ forms washers when neither boundary reaches the axis. Measure both radii perpendicular to the axis. The outer radius $R$ is the farther distance and the inner radius $r$ is the nearer distance, regardless of which function is named first.

V=\pi\int\big(R^2-r^2\big),d(\text{perpendicular variable}),\qquad R=\text{farther distance to the axis},\quad r=\text{nearer distance}

Use slices perpendicular to the rotation line: $dx$ for a horizontal axis, $dy$ for a vertical axis.\n2. Write each boundary's distance from the line.\n3. Compare those distances and label the larger one $R$ .\n4. Integrate $\pi(R^2-r^2)$ over the matching bounds.\n5. Split the interval if the farther boundary changes.

Example: rotate the region between $y=x^2$ and $y=x$ on $0\le x\le1$ around $y=2$ . The axis lies above both curves. The lower curve $y=x^2$ is farther away, so $R(x)=2-x^2$ ; the upper curve is nearer, so $r(x)=2-x$ . Thus $V=\pi\int_0^1\left[(2-x^2)^2-(2-x)^2\right]dx=\pi\int_0^1(4x-5x^2+x^4)\,dx=\frac{8\pi}{15}$ cubic units.

“Upper” does not always mean “outer.” With an axis above the region, the lower boundary is farther from the axis; with an axis below, the upper boundary may be farther. Determine distance first, then square, and keep the area as $\pi(R^2-r^2)$ rather than $\pi(R-r)^2$ .

8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled

A short piece of a smooth graph has horizontal change $dx$ and vertical change $dy$ , so its length behaves like the hypotenuse $ds=\sqrt{dx^2+dy^2}$ . Dividing by the chosen differential produces a nonnegative length rate that can be accumulated across the interval.

y=f(x):\ L=\int_a^b\sqrt{1+[f'(x)]^2},dx;\qquad x=g(y):\ L=\int_c^d\sqrt{1+[g'(y)]^2},dy

Express the curve as one differentiable function of the integration variable.\n2. Differentiate that function.\n3. Form $\sqrt{1+(\text{derivative})^2}$ .\n4. Integrate over the coordinate interval.\n5. Check that the result is at least the straight-line distance between the endpoints.

Example: for $y=\frac23x^{3/2}$ on $0\le x\le1$ , $f'(x)=\sqrt{x}$ . Therefore $L=\int_0^1\sqrt{1+x}\,dx=\left[\frac23(1+x)^{3/2}\right]_0^1=\frac23(2\sqrt2-1)$ units.

The integrand contains the square of the derivative, not the square of the original function. Use this single-function formula only where the chosen orientation represents the curve smoothly across the full interval; otherwise a different orientation or a split into smooth pieces is needed.

Unit 8: Applications of Integration

Exam analysis

Most tested topics

8.1 Finding the Average Value of a Function on an Interval

Average Value Is Accumulation per Unit Interval

8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Integrals Accumulate Motion

8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts

An Accumulation Function Adds Up a Rate

Initial Amount Plus Integrated Net Rate Gives Final Amount

8.4 Finding the Area Between Curves Expressed as Functions of x

Vertical Slices Give Upper Minus Lower

8.5 Finding the Area Between Curves Expressed as Functions of y

Horizontal Slices Give Right Minus Left

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

Split Area Wherever the Curves Exchange Order

8.7 Volumes with Cross Sections: Squares and Rectangles

Build Volume from Square or Rectangular Slices

8.8 Volumes with Cross Sections: Triangles and Semicircles

Turn a Base Segment into a Geometric Cross Section

8.9 Volume with Disc Method: Revolving Around the x- or y-Axis

Build a Solid of Revolution from Discs

8.10 Volume with Disc Method: Revolving Around Other Axes

Measure Disc Radius from a Shifted Axis

8.11 Volume with Washer Method: Revolving Around the x- or y-Axis

Subtract the Hole with the Washer Method

8.12 Volume with Washer Method: Revolving Around Other Axes

Measure Both Washer Radii from the Rotation Line

8.13 The Arc Length of a Smooth, Planar Curve and Distance Traveled

Accumulate Tiny Distances to Find Arc Length