Areas Under Curves and Between Curves — Worked Solutions

Ques­tion 1: Area under a parabola#

The prob­lem#

Find the area under $y = 3x^2$ from $x = 1$ to $x = 2$.

Under­stand­ing the prob­lem#

"Area under" means the region between the curve $y = 3x^2$, the $x$-axis, and the ver­ti­cal lines $x = 1$ and $x = 2$.

The idea#

On $[1, 2]$, $3x^2 > 0$, so the curve is above the axis and the area is sim­ply $\displaystyle A = \int_1^2 y\,dx$.

Step-by-step solu­tion#

Step 1. Check the sign: for $1 \le x \le 2$, $3x^2$ is pos­i­tive, so no part of the region is below the axis.

Step 2. Set up the inte­gral.

$$\displaystyle A = \int_1^2 3x^2\,dx$$

Step 3. Inte­grate: an anti­deriv­a­tive of $3x^2$ is $x^3$.

$$A = \left[x^3\right]_1^2$$

Step 4. Sub­sti­tute the lim­its (upper minus lower).

$$A = 2^3 - 1^3 = 8 - 1 = 7$$

Check­ing the answer#

The curve rises from $3$ at $x = 1$ to $12$ at $x = 2$. A strip of width $1$ with height between $3$ and $12$ should have area between $3$ and $12$. $7$ fits.

Answer#

Area $= 7$ square units.

Ques­tion 2: A curve that crosses the axis#

The prob­lem#

Find the area bounded by $y = x^3$, the $x$-axis and $x = -1$, $x = 2$.

Under­stand­ing the prob­lem#

The cubic $y = x^3$ is neg­a­tive for $x < 0$ and pos­i­tive for $x > 0$. So between $x = -1$ and $x = 0$ the region lies below the $x$-axis, and between $x = 0$ and $x = 2$ it lies above.

The idea#

The les­son says: if the curve crosses the axis, split the inter­val at the cross­ing and take the absolute value of the part below. A sin­gle inte­gral from $-1$ to $2$ would let the neg­a­tive part can­cel some of the pos­i­tive part, like Exam­ple 2.

Step-by-step solu­tion#

Step 1. Find where the curve crosses the axis: $x^3 = 0$ gives $x = 0$, which lies inside $[-1, 2]$. Split there.

$$\displaystyle A = \left\lvert \int_{-1}^{0} x^3\,dx \right\rvert + \int_{0}^{2} x^3\,dx$$

Step 2. Eval­u­ate the part below the axis.

$$\displaystyle \int_{-1}^{0} x^3\,dx = \left[\frac{x^4}{4}\right]_{-1}^{0} = 0 - \frac{1}{4} = -\frac{1}{4}, \qquad \left\lvert -\frac{1}{4} \right\rvert = \frac{1}{4}$$

Step 3. Eval­u­ate the part above the axis.

$$\displaystyle \int_{0}^{2} x^3\,dx = \left[\frac{x^4}{4}\right]_{0}^{2} = \frac{16}{4} - 0 = 4$$

Step 4. Add the two areas.

$$\displaystyle A = \frac{1}{4} + 4 = \frac{17}{4}$$

Check­ing the answer#

The plain inte­gral $\displaystyle \int_{-1}^{2} x^3\,dx = 4 - \tfrac{1}{4} = \tfrac{15}{4}$ is smaller than our answer by $\displaystyle 2 \times \tfrac{1}{4}$, exactly the can­cel­la­tion we avoided.

Answer#

Area $\displaystyle = \tfrac{17}{4}$ square units.

Com­mon mis­take to avoid#

Writ­ing $\displaystyle \int_{-1}^{2} x^3\,dx = \tfrac{15}{4}$ as the area. Area is never reduced by parts below the axis.

Ques­tion 3: Region inside a parabola cut off by a line#

The prob­lem#

Find the area enclosed by $y^2 = 8x$ and the line $x = 2$.

Under­stand­ing the prob­lem#

$y^2 = 8x$ is a parabola with ver­tex at the ori­gin, open­ing to the right, sym­met­ric about the $x$-axis. The ver­ti­cal line $x = 2$ cuts it at $y^2 = 16$, i.e. at $(2, 4)$ and $(2, -4)$. The enclosed region is the "cap" from the ver­tex out to $x = 2$.

The idea#

Use sym­me­try, as in Exam­ple 3: find the area of the upper half (between $y = \sqrt{8x}$ and the $x$-axis from $x = 0$ to $x = 2$) and dou­ble it.

Step-by-step solu­tion#

Step 1. Write the upper half of the parabola as a func­tion. $y^2 = 8x$ gives, for $y \ge 0$,

$$y = \sqrt{8x} = 2\sqrt{2}\,\sqrt{x}$$

Step 2. Set up the area as twice the upper half.

$$\displaystyle A = 2\int_0^2 2\sqrt{2}\,x^{1/2}\,dx = 4\sqrt{2}\int_0^2 x^{1/2}\,dx$$

Step 3. Inte­grate: $\displaystyle \int x^{1/2}\,dx = \tfrac{2}{3}x^{3/2}$.

$$\displaystyle A = 4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^2 = 4\sqrt{2} \cdot \frac{2}{3} \cdot 2^{3/2}$$

Step 4. Sim­plify. $2^{3/2} = 2\sqrt{2}$, so $\sqrt{2} \cdot 2\sqrt{2} = 4$.

$$\displaystyle A = 4 \cdot \frac{2}{3} \cdot 4 = \frac{32}{3}$$

Check­ing the answer#

The region fits inside the rec­tan­gle $0 \le x \le 2$, $-4 \le y \le 4$, of area $16$. A par­a­bolic cap is known to fill $\displaystyle \tfrac{2}{3}$ of its bound­ing rec­tan­gle: $\displaystyle \tfrac{2}{3} \times 16 = \tfrac{32}{3}$. It agrees.

Answer#

Area $\displaystyle = \tfrac{32}{3}$ square units.

Ques­tion 4: Area of an ellipse#

The prob­lem#

Find the area of the ellipse $\displaystyle \frac{x^2}{25} + \frac{y^2}{9} = 1$.

Under­stand­ing the prob­lem#

This ellipse is cen­tred at the ori­gin, with $a^2 = 25$ and $b^2 = 9$, so $a = 5$ (half-width along $x$) and $b = 3$ (half-height along $y$). It is sym­met­ric in both axes.

The idea#

Find the area in the first quad­rant and mul­ti­ply by $4$, as in Exam­ple 4. Solve the equa­tion for $y$ to get the upper half, and use the stan­dard result $\displaystyle \int_0^a \sqrt{a^2 - x^2}\,dx = \tfrac{\pi a^2}{4}$ (the area of a quar­ter cir­cle of radius $a$).

Step-by-step solu­tion#

Step 1. Solve for $y \ge 0$.

$$\displaystyle \frac{y^2}{9} = 1 - \frac{x^2}{25} = \frac{25 - x^2}{25} \;\Longrightarrow\; y = \frac{3}{5}\sqrt{25 - x^2}$$

Step 2. Set up four times the first-quad­rant area. In the first quad­rant $x$ runs from $0$ to $5$.

$$\displaystyle A = 4\int_0^5 \frac{3}{5}\sqrt{25 - x^2}\,dx = \frac{12}{5}\int_0^5 \sqrt{25 - x^2}\,dx$$

Step 3. Eval­u­ate the inte­gral with the for­mula $\displaystyle \int \sqrt{a^2 - x^2}\,dx = \tfrac{x}{2}\sqrt{a^2 - x^2} + \tfrac{a^2}{2}\sin^{-1}\tfrac{x}{a}$, with $a = 5$.

$$\displaystyle \int_0^5 \sqrt{25 - x^2}\,dx = \left[\frac{x}{2}\sqrt{25 - x^2} + \frac{25}{2}\sin^{-1}\frac{x}{5}\right]_0^5 = \left(0 + \frac{25}{2}\cdot\frac{\pi}{2}\right) - 0 = \frac{25\pi}{4}$$

Step 4. Mul­ti­ply.

$$\displaystyle A = \frac{12}{5} \cdot \frac{25\pi}{4} = 15\pi$$

Check­ing the answer#

The gen­eral result is $\pi ab = \pi \times 5 \times 3 = 15\pi$. If $a = b$, the ellipse would be a cir­cle with area $\pi r^2$, so the for­mula behaves sen­si­bly.

Answer#

Area $= 15\pi$ square units.

Ques­tion 5: Area under a cosine curve#

The prob­lem#

Find the area bounded by $y = \cos x$ and the $x$-axis for $0 \le x \le \pi$.

Under­stand­ing the prob­lem#

$\cos x$ is pos­i­tive from $0$ to $\displaystyle \tfrac{\pi}{2}$ and neg­a­tive from $\displaystyle \tfrac{\pi}{2}$ to $\pi$. So the region has a part above the axis and a part below.

The idea#

Split at the cross­ing $\displaystyle x = \tfrac{\pi}{2}$ and take the absolute value of the part below, exactly as Exam­ple 2 does for $\sin x$.

Step-by-step solu­tion#

Step 1. Find the cross­ing: $\cos x = 0$ at $\displaystyle x = \tfrac{\pi}{2}$ in $[0, \pi]$.

Step 2. Area of the part above the axis.

$$\displaystyle \int_0^{\pi/2} \cos x\,dx = \left[\sin x\right]_0^{\pi/2} = 1 - 0 = 1$$

Step 3. Area of the part below the axis.

$$\displaystyle \int_{\pi/2}^{\pi} \cos x\,dx = \left[\sin x\right]_{\pi/2}^{\pi} = 0 - 1 = -1, \qquad \lvert -1 \rvert = 1$$

Step 4. Add.

$$A = 1 + 1 = 2$$

Check­ing the answer#

The graph of $\cos x$ on $[0, \pi]$ is sym­met­ric about the point $\displaystyle \left(\tfrac{\pi}{2}, 0\right)$, so the two pieces must have equal areas. They do: $1$ each. The plain inte­gral $\displaystyle \int_0^\pi \cos x\,dx = 0$ shows why split­ting is essen­tial.

Answer#

Area $= 2$ square units.

Ques­tion 6: Area between a line and a parabola#

The prob­lem#

Find the area between $y = x$ and $y = x^2$.

Under­stand­ing the prob­lem#

The line $y = x$ and the parabola $y = x^2$ meet at two points and enclose a small region between them. No lim­its are given, so you must find them.

The idea#

Find the inter­sec­tion points; between them, decide which curve is on top; then inte­grate (top $-$ bot­tom), as in Exam­ple 6.

Step-by-step solu­tion#

Step 1. Find the inter­sec­tions.

$$x^2 = x \;\Longrightarrow\; x^2 - x = 0 \;\Longrightarrow\; x(x - 1) = 0 \;\Longrightarrow\; x = 0 \text{ or } x = 1$$

Step 2. Decide which is on top. Test $\displaystyle x = \tfrac{1}{2}$: the line gives $\displaystyle \tfrac{1}{2}$, the parabola gives $\displaystyle \tfrac{1}{4}$. So $y = x$ is on top for $0 < x < 1$.

Step 3. Set up and inte­grate.

$$\displaystyle A = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3}$$

Step 4. Sim­plify.

$$\displaystyle A = \frac{3 - 2}{6} = \frac{1}{6}$$

Check­ing the answer#

The region fits inside the tri­an­gle under $y = x$ from $0$ to $1$, whose area is $\displaystyle \tfrac{1}{2}$. $\displaystyle \tfrac{1}{6}$ is pos­i­tive and smaller than $\displaystyle \tfrac{1}{2}$.

Answer#

Area $\displaystyle = \tfrac{1}{6}$ square unit.

Com­mon mis­take to avoid#

Inte­grat­ing $(x^2 - x)$ and get­ting $\displaystyle -\tfrac{1}{6}$. Always put the upper curve first.

Ques­tion 7: Parabola and the x-axis#

The prob­lem#

Find the area between the parabola $y = 4 - x^2$ and the $x$-axis.

Under­stand­ing the prob­lem#

$y = 4 - x^2$ is a down­ward-open­ing parabola with ver­tex $(0, 4)$. It meets the $x$-axis where $y = 0$. The region is the "arch" above the axis between those two points.

The idea#

Find the $x$-inter­cepts to get the lim­its, then inte­grate $y$ between them (the curve is above the axis there).

Step-by-step solu­tion#

Step 1. Find the $x$-inter­cepts.

$$4 - x^2 = 0 \;\Longrightarrow\; x^2 = 4 \;\Longrightarrow\; x = -2 \text{ or } x = 2$$

Step 2. Between $-2$ and $2$, $4 - x^2 \ge 0$, so the area is

$$\displaystyle A = \int_{-2}^{2} (4 - x^2)\,dx$$

Step 3. Inte­grate.

$$\displaystyle A = \left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right)$$

Step 4. Sim­plify.

$$\displaystyle A = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$$

Check­ing the answer#

The arch sits inside a rec­tan­gle $4$ wide and $4$ tall (area $16$), and a par­a­bolic arch fills $\displaystyle \tfrac{2}{3}$ of its rec­tan­gle: $\displaystyle \tfrac{2}{3} \times 16 = \tfrac{32}{3}$. It agrees.

Answer#

Area $\displaystyle = \tfrac{32}{3}$ square units.

Ques­tion 8: Parabola and a slanted line#

The prob­lem#

Find the area bounded by $y = x^2$ and the line $y = x + 2$.

Under­stand­ing the prob­lem#

The upward parabola $y = x^2$ and the line $y = x + 2$ cross at two points. The bounded region is the part below the line and above the parabola.

The idea#

Find the inter­sec­tions, check which curve is on top, and inte­grate (line $-$ parabola).

Step-by-step solu­tion#

Step 1. Find the inter­sec­tions.

$$x^2 = x + 2 \;\Longrightarrow\; x^2 - x - 2 = 0 \;\Longrightarrow\; (x - 2)(x + 1) = 0$$

So $x = -1$ (point $(-1, 1)$) and $x = 2$ (point $(2, 4)$).

Step 2. Check which is on top. At $x = 0$: line gives $2$, parabola gives $0$. So the line is on top.

Step 3. Set up and inte­grate.

$$\displaystyle A = \int_{-1}^{2} (x + 2 - x^2)\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}$$

Step 4. Sub­sti­tute the lim­its.

$$\displaystyle \begin{aligned} \text{at } x = 2: &\quad 2 + 4 - \frac{8}{3} = \frac{10}{3} \\ \text{at } x = -1: &\quad \frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6} \end{aligned}$$

Step 5. Sub­tract.

$$\displaystyle A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$$

Check­ing the answer#

For a parabola $y = x^2$ cut by a line meet­ing it at $x = \alpha$ and $x = \beta$, the enclosed area is $\displaystyle \tfrac{(\beta - \alpha)^3}{6}$. Here $\displaystyle \tfrac{3^3}{6} = \tfrac{27}{6} = \tfrac{9}{2}$. It agrees.

Answer#

Area $\displaystyle = \tfrac{9}{2}$ square units.

Ques­tion 9: A seg­ment cut from a cir­cle#

The prob­lem#

Find the area of the smaller region cut from the cir­cle $x^2 + y^2 = 16$ by the line $x = 2$.

Under­stand­ing the prob­lem#

The cir­cle has cen­tre $(0, 0)$ and radius $4$. The ver­ti­cal line $x = 2$ cuts it into two pieces. The smaller piece is to the right of the line, run­ning from $x = 2$ to the edge of the cir­cle at $x = 4$. It is sym­met­ric about the $x$-axis.

The idea#

Find the area of the upper half of that piece (between $y = \sqrt{16 - x^2}$ and the $x$-axis from $x = 2$ to $x = 4$) and dou­ble it. Use the stan­dard inte­gral

$$\displaystyle \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C.$$

Step-by-step solu­tion#

Step 1. Write the upper semi­cir­cle: $y = \sqrt{16 - x^2}$, and set up the area.

$$\displaystyle A = 2\int_2^4 \sqrt{16 - x^2}\,dx$$

Step 2. Write the anti­deriv­a­tive with $a = 4$ ($\displaystyle \tfrac{a^2}{2} = 8$).

$$\displaystyle F(x) = \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}$$

Step 3. Eval­u­ate at the upper limit $x = 4$.

$$\displaystyle F(4) = 2 \cdot 0 + 8\sin^{-1}1 = 8 \cdot \frac{\pi}{2} = 4\pi$$

Step 4. Eval­u­ate at the lower limit $x = 2$. Here $\sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$ and $\displaystyle \sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6}$.

$$\displaystyle F(2) = 1 \cdot 2\sqrt{3} + 8 \cdot \frac{\pi}{6} = 2\sqrt{3} + \frac{4\pi}{3}$$

Step 5. Sub­tract, then dou­ble.

$$\displaystyle \int_2^4 \sqrt{16 - x^2}\,dx = 4\pi - \frac{4\pi}{3} - 2\sqrt{3} = \frac{8\pi}{3} - 2\sqrt{3}$$

$$\displaystyle A = 2\left(\frac{8\pi}{3} - 2\sqrt{3}\right) = \frac{16\pi}{3} - 4\sqrt{3}$$

Check­ing the answer#

Numer­i­cally, $\displaystyle \tfrac{16\pi}{3} - 4\sqrt{3} \approx 16.76 - 6.93 = 9.83$. The whole cir­cle has area $16\pi \approx 50.3$, so the smaller piece is about a fifth of it, which matches a line cut­ting halfway between the cen­tre and the edge. Geo­met­ri­cally, it is a sec­tor of angle $\displaystyle \tfrac{2\pi}{3}$ (area $\displaystyle \tfrac{16\pi}{3}$) minus a tri­an­gle of area $\displaystyle \tfrac{1}{2} \cdot 4\sqrt{3} \cdot 2 = 4\sqrt{3}$. It agrees.

Answer#

Area $\displaystyle = \frac{16\pi}{3} - 4\sqrt{3}$ square units ($\approx 9.83$).

Ques­tion 10: Area of a tri­an­gle by inte­gra­tion#

The prob­lem#

Find the area of the tri­an­gle with ver­tices $(0, 0)$, $(4, 0)$, $(2, 3)$ by inte­gra­tion.

Under­stand­ing the prob­lem#

The base runs along the $x$-axis from $(0, 0)$ to $(4, 0)$; the top ver­tex is $(2, 3)$. The upper bound­ary is made of two dif­fer­ent lines: from $(0, 0)$ up to $(2, 3)$, then from $(2, 3)$ down to $(4, 0)$. You must use inte­gra­tion, not the $\displaystyle \tfrac{1}{2} \times$ base $\times$ height for­mula (though you may use it to check).

The idea#

Find the equa­tion of each slanted side, then inte­grate each over its own inter­val and add. The split point is $x = 2$, where the upper bound­ary changes from one line to the other.

Step-by-step solu­tion#

Step 1. Line through $(0, 0)$ and $(2, 3)$: slope $\displaystyle \tfrac{3 - 0}{2 - 0} = \tfrac{3}{2}$, through the ori­gin.

$$\displaystyle y = \frac{3x}{2}$$

Step 2. Line through $(2, 3)$ and $(4, 0)$: slope $\displaystyle \tfrac{0 - 3}{4 - 2} = -\tfrac{3}{2}$. Using the point $(4, 0)$: $\displaystyle y = -\tfrac{3}{2}(x - 4)$.

$$\displaystyle y = 6 - \frac{3x}{2}$$

Step 3. Area of the left part, from $x = 0$ to $x = 2$.

$$\displaystyle \int_0^2 \frac{3x}{2}\,dx = \left[\frac{3x^2}{4}\right]_0^2 = \frac{12}{4} = 3$$

Step 4. Area of the right part, from $x = 2$ to $x = 4$.

$$\displaystyle \int_2^4 \left(6 - \frac{3x}{2}\right)dx = \left[6x - \frac{3x^2}{4}\right]_2^4 = (24 - 12) - (12 - 3) = 12 - 9 = 3$$

Step 5. Add.

$$A = 3 + 3 = 6$$

Check­ing the answer#

$\displaystyle \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 4 \times 3 = 6$. The inte­gra­tion agrees with geom­e­try.

Answer#

Area $= 6$ square units.