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Math 220-2: Kursobjekt

Aaron Greicius

Section 1.8 More substitution; area between curves

Before discussing the area of a region lying between two curves, we provide an alternative method to Procedure 1.7.6for definite integral substitution. We encourage you to try your hand at both methods; you may find that different circumstances will make one method more convenient than the other.

Example 1.8.2. Definite integral substitution: streamlined.

Use streamlined substitution technique to compute the following integrals.
  1. \(\displaystyle \displaystyle\int_{\pi}^{2\pi}\cos^2(x)\sin x\, dx\)
  2. \(\displaystyle \displaystyle\int_{1}^{2} \sqrt{s^8+s^6}\, ds\)
Solution.
  1. Using the substitution
    \begin{align*} u \amp =\cos x\\ du \amp = -\sin x\text{,} \end{align*}
    the streamlined technique yields
    \begin{align*} \int_\pi^{2\pi}\cos^2x\sin x\, dx \amp = \int_{u=\cos \pi}^{u=\cos 2\pi}-u^2\, du\\ \amp = -\frac{1}{3}u^3\Bigr\vert_{-1}^{1}\\ \amp = -\frac{2}{3}\text{.} \end{align*}
  2. Before substituting, we do a bit of algebra:
    \begin{equation*} \sqrt{s^8+s^6}=\sqrt{s^6(s^2+1)}=s^3\sqrt{s^2+1}\text{.} \end{equation*}
    This suggests the substitution
    \begin{align*} u \amp = s^2+1\\ du \amp =2s\, ds \text{.} \end{align*}
    To transform the given integral in terms of \(u\text{,}\) we further observe that
    \begin{equation*} s^3\, ds= s^2\cdot s\, ds=\frac{1}{2}(u-1)du\text{.} \end{equation*}
    Now using the streamlined method, we have
    \begin{align*} \int_{1}^{2} \sqrt{s^8+s^6}\, ds \amp =\int_{1}^{2} s^3\sqrt{s^2+1}\, ds \\ \amp = \int_{u=1^2+1}^{u=2^2+1}\frac{1}{2}(u-1)\sqrt{u}\, du\\ \amp = \frac{1}{2}\int_2^5 u^{3/2}-u^{1/2} \\ \amp = \frac{1}{2}\left( \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right)\Bigr\vert_1^5\\ \amp = \frac{10}{3}\sqrt{5}-\frac{2}{15}\sqrt{2}\text{.} \end{align*}

Definition 1.8.3. Area between curves: functions of \(x\).

Suppose \(f(x)\geq g(x)\) for all \(x\in [a,b]\text{.}\) Let \(\mathcal{R}\) be the region between the graph of \(f\) and the graph of \(g\text{,}\) and between the vertical lines \(x=a\) and \(x=b\text{:}\) i.e.,
\begin{equation*} \mathcal{R}=\{(x,y)\colon a\leq x\leq b, g(x)\leq y\leq f(x)\}\text{.} \end{equation*}
We define the area of \(\mathcal{R}\) to be the integral of \(f-g\) over \([a,b]\text{:}\) i.e.,
\begin{equation*} \text{ area } (\mathcal{R})=\int_a^b f(x)-g(x)\, dx\text{.} \end{equation*}

Definition 1.8.4. Area between curves: functions of \(y\).

Suppose \(x=p(y)\) and \(x=q(y)\) are two functions of \(y\) satisfying \(p(y)\geq q(y)\) for all \(y\in [c,d]\text{.}\) Let \(\mathcal{R}\) be the region between the graph of \(p\) and the graph of \(q\text{,}\) and between the horizontal lines \(y=c\) and \(y=d\text{:}\) i.e.,
\begin{equation*} \mathcal{R}=\{(x,y)\colon c\leq y\leq d, p(y)\leq x\leq q(y)\}\text{.} \end{equation*}
We define the area of \(\mathcal{R}\) to be the integral of \(p-q\) over \([c,d]\text{:}\) i.e.,
\begin{equation*} \text{ area } (\mathcal{R})=\int_c^d p(y)-q(y)\, dy\text{.} \end{equation*}

Remark 1.8.5. Area between curves: why \(f-g\text{?}\)

Suppose \(f(x)\geq g(x)\) for all \(x\in [a,b]\text{.}\) Let \(\mathcal{C}_1\) be the graph of \(f\text{,}\) let \(\mathcal{C}_2\) be the graph of \(g\text{,}\) and \(\mathcal{R}\) be the region between \(\mathcal{C}_1\) and \(\mathcal{C}_2\) over the interval \([a,b]\) on the \(x\)-axis.
  1. Suppose we also have \(f(x)\geq g(x)\geq 0\) for all \(x\in [a,b]\text{.}\) Then we have
    \begin{align*} \text{ area } (\mathcal{R})\amp =\int_a^b f(x)-g(x)\, dx\\ \amp =\int_a^b f(x)\, dx -\int_a^bg(x)\, dx\\ \amp =\text{ area } (\mathcal{R}_1)-\text{ area } (\mathcal{R}_2)\text{,} \end{align*}
    where \(\mathcal{R}_i\) is the region lying between \(\mathcal{C}_i\) and the \(x\)-axis over the interval \([a,b]\text{.}\)
    Area as difference of two areas
  2. To reduce the general case \(f(x)\geq g(x)\) to the case above, simply shift both functions (and hence also \(\mathcal{R}\)) up by a large enough constant \(C\) so that \(f(x)\geq g(x)\geq 0\text{.}\) This operation does not affect the area of \(\mathcal{R}\text{,}\) and the \(C\) gets canceled in the integral computation thanks to the difference operator!
    Shifting the region up

Example 1.8.6. Area between parabola and line.

Let \(\mathcal{R}\) be the region between the parabola \(x+y^2=4\) and the line \(2x+y=2\) lying in the first quadrant. Compute the are of \(\mathcal{R}\text{.}\) You may do this either by thinking of the curves as graphs of functions of \(x\text{,}\) or graphs of functions of \(y\text{.}\) Which approach is easier?
Solution.
Below you find a diagram of \(\mathcal{R}\text{.}\) The setup lends itself to a treatment in terms of \(y\) in part because for both equations defining the curves we can solve for \(x\) easily in terms of \(y\text{.}\) The region is naturally described as the points lying between the curves
\begin{align*} x=p(y) \amp = 4-y^2\\ x=q(y) \amp= 1-y/2 \end{align*}
and between the lines \(y=0\) and \(y=2\text{.}\) Since \(p(y)\geq q(y)\) for all \(y\in [0,2]\) we may use the area formula to compute
\begin{align*} \operatorname{area}\mathcal{R} \amp = \int_0^2 p(y)-q(y)\, dy\\ \amp = \int_0^2 3-y^2+y/2\, dy\\ \amp = 13/3\text{.} \end{align*}
Region between parabola and line

Example 1.8.8. Area between parabolas.

Compute the area of the region between the parabolas \(y=-x^2-2x\) and \(y=x^2-4\) lying within the lines \(x=-3\) and \(x=2\text{.}\)
Solution.
In the diagram below the region \(\mathcal{R}\) is described as a union of three regions \(\mathcal{R}_i\text{,}\) each of whose area can be computed with the area formula. We conclude
\begin{align*} \operatorname{area}\mathcal{R} \amp=\operatorname{area}\mathcal{R}_1+\operatorname{area}\mathcal{R}_2+\operatorname{area}\mathcal{R}_3 \\ \amp = \int_{-3}^{-2}(x^2-4)-(-x^2-2x)\, dx + \int_{-2}^{1}(-x^2-2x)-(x^2-4)\, dx + \int_{1}^{2}(x^2-4)-(-x^2-2x)\, dx\\ \amp =\frac{11}{3}+9+\frac{11}{3}\\ \amp = \frac{49}{3} \end{align*}
.
Region between two parabolas
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