Application of Integrals

1. Check this before you revise anything

"Area between two curves" does not exist in this edition. Older editions of this chapter had two sections: 8.2 Area under Simple Curves, and a second section on the area enclosed between two curves — find their intersection points, integrate [upper curve − lower curve].

The current book has exactly one section, 8.2 Area under Simple Curves, and nothing else. A full-text search of the entire 8-page chapter for "two curves," "upper minus lower," or any intersection-point technique returns zero hits.

This matches the syllabus line exactly: "Applications in finding the area under simple curves, especially lines, circles/parabolas/ellipses (in standard form only)." Nothing about a second curve.

The old stub taught area-between-two-curves as its own formula, its own procedure ("find intersection points, determine which curve is upper, integrate the difference"), and called it the "MOST COMMON EXAM TYPE." That entire technique — genuinely useful in other syllabi — is not part of this book's current chapter and is removed from this rebuild.

This is also the smallest chapter in the entire book. One exercise (Exercise 8.1, 4 questions: 2 ellipses, 2 MCQs) plus a Miscellaneous Exercise (5 questions, one with two parts) — 9 questions in total, against 261 in the previous chapter alone. The old stub invented a much larger structure with no real exercises or solutions file behind it.

2. Where the Area Formula Comes From (Textbook 8.2)

The previous chapter established the definite integral as the limit of a sum. This chapter turns that into a tool for measuring regions whose boundaries are curved, which no formula from elementary geometry can handle.

The elementary strip. Consider the region bounded by the curve $y=f(x)$, the x-axis, and the ordinates $x=a$ and $x=b$. Think of it as made up of a very large number of very thin vertical strips.

Take one arbitrary strip at position $x$, of height $y$ and width $dx$. Its area — the elementary area — is:

$dA=ydx\text{where }y=f(x)$

The word "arbitrary" is doing real work here. The strip is not at any particular place; it stands for every strip, located at some unspecified $x$ between $a$ and $b$. That is what allows the sum to become an integral.

Adding the strips. The total area is the result of adding up these elementary areas right across the region, which the book writes symbolically as:

$A={\int }_a^{b}dA={\int }_a^{b}ydx={\int }_a^{b}f(x)dx$

Horizontal strips. The same argument runs sideways. For the region bounded by the curve $x=g(y)$, the y-axis, and the lines $y=c$ and $y=d$, take horizontal strips of width $dy$:

$A={\int }_c^{d}xdy={\int }_c^{d}g(y)dy$

Choosing between vertical and horizontal strips is a matter of which one makes the integral easier. Both give the same answer, and the textbook demonstrates this explicitly by computing the area of a circle each way in Example 1.

3. Curves That Cross the Axis (Textbook 8.2, Remark)

This is where the geometric area and the signed integral part company, and it is the most commonly examined subtlety in the chapter.

Entirely below the axis. If the curve lies below the x-axis on $[a,b]$, then $f(x)<0$ throughout, and the integral ${\int }_a^{b}f(x)dx$ comes out negative. An area cannot be negative, so only the numerical value is taken:

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

Partly above, partly below. More often, part of the curve is above the axis and part below. Suppose the region splits into a piece with signed area $A_1<0$ and a piece with $A_2>0$. The geometric area is:

$A=|A_1|+A_2$

Why the naive integral fails. If you integrate straight across without splitting, the negative and positive contributions cancel, and you get an answer that is too small — sometimes zero.

The procedure is therefore always the same: find where $f(x)=0$ inside the interval, split the integral at those points, take the absolute value of each piece, and add.

Worked, mirroring the textbook's own technique. Find the area bounded by $y=\sin x$ between $x=0$ and $x=2\pi$.

Since $\sin x\ge 0$ on $[0,\pi ]$ and $\sin x\le 0$ on $[\pi ,2\pi ]$, split at $x=\pi$:

$A={\int }_0^{\pi }\sin xdx+\left|{\int }_{\pi }^{2\pi }\sin xdx\right|=2+|-2|=4$

The naive single integral ${\int }_0^{2\pi }\sin xdx$ gives $0$ — the two halves cancel exactly. That is a correct signed integral and a wrong area.

The book makes the same point with the line $y=3x+2$ on $[-1,1]$, which crosses the axis at $x=-\frac{2}{3}$, requiring the interval to be split there.

4. Circles and Ellipses in Standard Position (Textbook 8.2, Examples 1 to 2)

Closed curves are handled by the same single-curve formula, with symmetry doing the work of reducing the problem to one quadrant.

The circle $x^2+y^2=a^2$. The circle is symmetric about both axes, so the whole enclosed area is four times the area of the first-quadrant region bounded by the curve, the x-axis, and the ordinates $x=0$ and $x=a$:

$A=4{\int }_0^{a}ydx=4{\int }_0^a\sqrt{a^2-x^2}dx$

Solving $x^2+y^2=a^2$ gives $y=\pm \sqrt{a^2-x^2}$, and the positive root is taken because the region lies in the first quadrant. Choosing the wrong root here is a standard error.

Using the standard result $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}$ from Chapter 7:

$A=4{\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right]}_0^a=4\cdot \frac{a^2}{2}\cdot \frac{\pi }{2}=\pi a^2$

which is the familiar formula, now derived rather than assumed.

The same area with horizontal strips. Integrating $4{\int }_0^{a}xdy=4{\int }_0^a\sqrt{a^2-y^2}dy$ produces $\pi a^2$ again. The textbook includes this deliberately, to show that the choice of strip direction is free.

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Identical method. By symmetry about both axes, take four times the first-quadrant region, where $y=\frac{b}{a}\sqrt{a^2-x^2}$:

$A=4{\int }_0^a\frac{b}{a}\sqrt{a^2-x^2}dx=\frac{4b}{a}\cdot \frac{\pi a^2}{4}=\pi ab$

The result $A=\pi ab$ holds regardless of which semi-axis is longer, so there is no need to check whether $a>b$ before using it. Setting $a=b$ recovers the circle.

The general recipe for every question in Exercise 8.1: identify the symmetry, restrict to one quadrant, solve the curve's equation for $y$ (taking the correct sign), integrate, and multiply back up.

Summary

• Area is built from elementary strips: a vertical strip at an arbitrary $x$ has area $dA=ydx$, and adding them across the region gives $A={\int }_a^{b}f(x)dx$.
• Horizontal strips give the mirror formula $A={\int }_c^{d}g(y)dy$ for a curve $x=g(y)$ against the y-axis; either direction is valid, so pick the easier integral.
• A curve below the axis makes the signed integral negative — take its absolute value, since only the numerical value counts as area.
• When the curve crosses the axis, split the interval at the zeros and add the absolute values, $A=|A_1|+A_2$; integrating straight across lets the pieces cancel and gives the wrong area.
• Circles and ellipses in standard position use the same single-curve formula plus symmetry: compute one quadrant and multiply by 4.
• Solve the curve for $y$ taking the sign appropriate to the quadrant — the positive root in the first quadrant.
• ${\int }_0^a\sqrt{a^2-x^2}dx$ gives the circle area $\pi a^2$; the ellipse gives $\pi ab$ regardless of which semi-axis is longer, and reduces to the circle when $a=b$.
• "Area between two curves" — finding intersection points and integrating [upper − lower] — is not part of the current edition of this chapter, despite being a full section in older editions.
• This is the smallest chapter in the book: one exercise, 9 questions in total.