Application of Integrals

The One Thing

Definite integration calculates the exact area of a region bounded by curves, straight lines, and coordinate axes. The essential procedure is to identify the region, determine all intercepts and intersections, choose vertical or horizontal strips, and integrate the appropriate upper-minus-lower or right-minus-left expression over the correct limits.

Definitions and Results

• Definite integral: A definite integral gives a numerical value over specified limits and is commonly used to calculate accumulated quantities such as area.

• Area under a curve: The area between $y=f(x)$, the $x$-axis, and the vertical lines $x=a$ and $x=b$ is

$$A={\int }_a^b|f(x)|dx.$$ If $f(x)$ is non-negative on $[a,b]$, then $$A={\int }_a^{b}f(x)dx.$$ If $f(x)$ is non-positive on $[a,b]$, then $$A=-{\int }_a^{b}f(x)dx.$$

• Area with respect to $x$: For vertical strips,

$$A={\int }_a^b[\text{upper function}-\text{lower function}]dx.$$ More generally, for $y=f(x)$ and $y=g(x)$, $$A={\int }_a^b|f(x)-g(x)|dx.$$ If $f(x)$ is above $g(x)$ throughout the interval, then $$A={\int }_a^b[f(x)-g(x)]dx.$$

• Area with respect to $y$: For horizontal strips,

$$A={\int }_c^d[\text{right function}-\text{left function}]dy.$$ If the boundaries are $x=\varphi (y)$ and $x=\psi (y)$, then $$A={\int }_c^d|\varphi (y)-\psi (y)|dy.$$

• Region bounded by coordinate axes: A closed region formed partly by the $x$-axis, $y$-axis, and one or more curves. For a first-quadrant region bounded by $y=f(x)$, the $x$-axis, and the $y$-axis, the usual limits are $x=0$ to the positive $x$-intercept.

• $x$-intercept: The point where a curve meets the $x$-axis, found by putting $y=0$, or equivalently solving $f(x)=0$.

• $y$-intercept: The point where a curve meets the $y$-axis, found by putting $x=0$, or evaluating $f(0)$.

• Points of intersection: Points common to two curves, found by solving their equations simultaneously.

• Signed area: A definite integral gives a positive contribution above the $x$-axis and a negative contribution below it. Geometric area must therefore use absolute values or split the interval where the sign changes.

• Symmetry: Symmetry about an axis can reduce the calculation. For a region symmetric about the $y$-axis,

$$A_{\text{total}}=2{\int }_0^{\text{positive boundary}}(\text{height})dx.$$

• Fundamental Theorem of Calculus: If $F^{\prime }(x)=f(x)$, then

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

• Common antiderivatives:

$$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1,$$ $$\int \frac{1}{x}dx=\ln |x|+C,$$ $$\int e^{x}dx=e^x+C.$$

• Area as a geometric quantity: The final answer must be non-negative and should be reported in square units.

Worked Methods

Method 1: Area under a single curve

• Sketch $y=f(x)$, the $x$-axis, and the vertical boundaries $x=a$ and $x=b$.
• Determine whether $f(x)$ is non-negative or non-positive on $[a,b]$.
• Use

$$A={\int }_a^b|f(x)|dx.$$

• If $f(x)\ge 0$, integrate $f(x)$ directly.
• If $f(x)\le 0$, use the negative of the integral.
• Evaluate the antiderivative using

$${\int }_a^{b}f(x)dx=F(b)-F(a).$$

• State the answer as a positive number of square units.

If the curve crosses the $x$-axis within $[a,b]$, find the crossing point by solving $f(x)=0$, split the integral there, and calculate each geometric area separately.

Method 2: Area between two curves using $dx$

• Sketch the curves $y=f(x)$ and $y=g(x)$.
• Find their points of intersection by solving

$$f(x)=g(x).$$

• Use the intersection $x$-coordinates as limits, or include any additional boundary limits specified by the region.
• Determine which curve is upper on each subinterval.
• Integrate vertical strip height:

$$A={\int }_a^b[\text{upper function}-\text{lower function}]dx.$$

• If the curves cross within the interval and change order, divide the integral at every point of intersection.
• Alternatively, use

$$A={\int }_a^b|f(x)-g(x)|dx.$$

Method 3: Area between two curves using $dy$

• Consider whether horizontal strips give a simpler description of the region.
• Express the boundaries as $x=\varphi (y)$ and $x=\psi (y)$, if necessary.
• Find the relevant $y$-coordinates from intercepts, intersections, or other boundaries.
• Determine which boundary is rightmost and which is leftmost.
• Integrate horizontal strip width:

$$A={\int }_c^d[\text{right function}-\text{left function}]dy.$$

• If the order changes, split the integral at the relevant $y$-value.
• Use

$$A={\int }_c^d|\varphi (y)-\psi (y)|dy$$ when the order is not constant.

A curve that is more conveniently expressed as $x=\varphi (y)$ may therefore be integrated with respect to $y$ rather than rewritten as $y=f(x)$.

Method 4: Region bounded by coordinate axes

• Sketch the curve and the coordinate axes.
• Find the $x$-intercept by solving $f(x)=0$.
• Find the $y$-intercept by evaluating $f(0)$.
• Identify the closed region enclosed by the curve, the $x$-axis, and the $y$-axis.
• For a first-quadrant region, use the limits $x=0$ to the positive $x$-intercept.
• Integrate the curve above the $x$-axis:

$$A={\int }_0^{\text{positive }x\text{-intercept}}f(x)dx.$$

• Report the result in square units.

Method 5: Using symmetry

• Confirm that the region is symmetric about the $y$-axis.
• Calculate the area on one side, usually from $x=0$ to the positive boundary.
• Double the result:

$$A_{\text{total}}=2A_{\text{one side}}.$$

• State the final result as a non-negative area.

Where It Goes Wrong

• Using ${\int }_a^{b}f(x)dx$ without checking whether the curve lies below the $x$-axis; geometric area requires a positive result.
• Subtracting the lower curve from the upper curve without first identifying which curve is upper throughout the interval.
• Failing to find all points of intersection before selecting the limits.
• Continuing with one integral after curves cross and change order; the interval must be split at each crossing.
• Using vertical strips when the region is more simply described by horizontal strips, or forgetting that $dx$ uses upper-minus-lower while $dy$ uses right-minus-left.
• Omitting the sketch, intercept conditions, or square units, even though these determine the correct enclosed region and final interpretation.

What Gets Asked

• Calculate the area under a specified curve between two vertical lines.
• Find the area bounded by a curve, the $x$-axis, and the coordinate axes.
• Determine $x$- and $y$-intercepts and use them as limits of integration.
• Find the area between two curves by integrating upper-minus-lower with respect to $x$.
• Find the area between two curves by integrating right-minus-left with respect to $y$.
• Solve simultaneous equations to locate points of intersection.
• Split an area calculation when a curve crosses the $x$-axis or when two curves change order.
• Use absolute values to convert a signed integral into geometric area.
• Exploit symmetry about the $y$-axis to halve the work.
• Evaluate definite integrals using the Fundamental Theorem of Calculus and standard antiderivatives.
• Interpret the final answer as a non-negative quantity in square units.