Chapter Summary and Practice

Problem
For positive $a$ and $b$, identify the region measured by $I=(a/b){\int }_0^b\sqrt{b^2-y^2}dy$ and evaluate it.

1. 1.The integration variable is $y$, so the strips are horizontal. Their length is $x=(a/b)\sqrt{b^2-y^2}$, with vertical positions from zero to $b$.
2. 2.Squaring and rearranging gives $x^2/a^2+y^2/b^2=1$. The square root is nonnegative, so this describes the right-hand boundary in the first quadrant.
3. 3.The integral measures one quadrant of the ellipse, not the whole ellipse. The standard square-root integral equals $\pi b^2/4$, so $I=(a/b)(\pi b^2/4)=\pi ab/4$.
4. 4.Only if the whole ellipse were requested would we multiply this answer by four. This interpretation prevents an unnoticed symmetry-factor error.

Problem
Find the area bounded by $y^2=12x$, the vertical axis, and $y=6$. Show why vertical and horizontal strips agree.

1. 1.With horizontal strips, the length is $x=y^2/12$ and the bounds are $0\le y\le 6$. Thus $A={\int }_0^6{y}^2/12dy=[y^3/36{]}_0^6=6$.
2. 2.At the horizontal cutoff, $y=6$ gives $x=3$. With vertical strips, the region lies between the upper branch $y=\sqrt{12x}$ and the line $y=6$, not between that branch and the axis.
3. 3.Thus $A={\int }_0^3(6-\sqrt{12x})dx$. The constant term contributes $18$, while the square-root term contributes $(2\sqrt{12}/3)3^{3/2}=12$.
4. 4.Subtracting gives $18-12=6$ square units again. The horizontal description is shorter, which is why it is the more efficient first choice here.

Problem
Find the first-quadrant area of $x^2+y^2=16$ and the full area of $x^2/16+y^2/4=1$. Explain why they differ.

1. 1.The circle has radius $4$, so its full area is $16\pi$. Only a quadrant is requested, giving $16\pi /4=4\pi$.
2. 2.The ellipse has semiaxes $4$ and $2$, so its full area is $8\pi$. Its first-quadrant area would be $2\pi$.
3. 3.The two requested regions are different fractions of different shapes. Comparing the equations alone without reading the requested region could introduce a factor-of-four error.
4. 4.An integral setup for the requested circular quadrant is ${\int }_0^4\sqrt{16-x^2}dx$. A setup for the full ellipse is $4{\int }_0^4(1/2)\sqrt{16-x^2}dx$, which is twice as large.

Problem
Find the geometric area between $y=-\sqrt{9-x^2}$ and the horizontal axis for $-3\le x\le 3$. Also state the signed integral.

1. 1.The negative square root is the lower half of the circle of radius $3$. The graph is nonpositive throughout the interval.
2. 2.Its geometric strip height is $\sqrt{9-x^2}$. The region is a semicircle, so its area is $\pi (3{)}^2/2=9\pi /2$ square units.
3. 3.The signed integral of the given negative function is $-9\pi /2$. Taking its magnitude is valid in this case because the sign stays nonpositive across the whole interval.
4. 4.The phrase lower semicircle chooses the branch; the word area chooses a nonnegative result. These are separate decisions.

Problem
For $y=2x-1$ on $[-1,2]$, find the signed integral and the geometric area.

1. 1.The intercept is $x=1/2$. The line is negative to its left and positive to its right. An antiderivative is $F(x)=x^2-x$.
2. 2.The endpoint values are $F(-1)=2$ and $F(2)=2$, so the whole signed integral is zero.
3. 3.At the intercept, $F(1/2)=-1/4$. The left signed contribution is $-1/4-2=-9/4$; the right is $2-(-1/4)=9/4$.
4. 4.The geometric area is $9/4+9/4=9/2$ square units. Two congruent triangles confirm the result: each has base $3/2$ and height $3$.

Problem
Find the area between $y=\sin x$ and the axis from zero to $3\pi$. Compare it with the signed integral.

1. 1.Use the zeros $0$, $\pi$, $2\pi$, and $3\pi$ as bounds. The three successive signed contributions are $2$, $-2$, and $2$.
2. 2.Their sum is the signed integral $2$. Their positive sizes sum to the area $2+2+2=6$ square units.
3. 3.Equivalently, each half-period has the same lobe area $2$. The interval contains three lobes, so the area is three times that of one lobe. Counting equal pieces works only after identifying the pieces correctly.

Problem
Let $p$ and $q$ be nonnegative. Find the area between $y=x^3$ and the axis from $x=-p$ to $x=q$, and compare it with the signed integral.

1. 1.For the negative-side piece, the signed integral is $[x^4/4{]}_{-p}^0=-p^4/4$. Its area is $p^4/4$.
2. 2.For the positive-side piece, the integral and area are both $[x^4/4{]}_0^q=q^4/4$.
3. 3.Thus $A=(p^4+q^4)/4$, whereas $S=(q^4-p^4)/4$. The sum and difference record the two different questions.
4. 4.If $p=q>0$, the signed integral is zero while the area is $p^4/2$. If one of the parameters is zero, the graph stays on one side and $A=|S|$.

Problem
Find ${\int }_0^3|x-1|dx$ and explain why no negative geometric contribution appears.

1. 1.The inside expression changes sign at one. Use $|x-1|=1-x$ on $[0,1]$ and $|x-1|=x-1$ on $[1,3]$.
2. 2.Calculate ${\int }_0^1(1-x)dx=1/2$ and ${\int }_1^3(x-1)dx=2$. Adding gives $5/2$.
3. 3.Both pieces are already nonnegative because they are values of an absolute-value function. The split is needed to write the correct formulas, not to reverse the sign of the function's contribution.