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Chapter 8 · Application of Integrals

Miscellaneous · Q1: Areas under y=x2y = x^2 on [1,2][1, 2] and y=x4y = x^4 on [1,5][1, 5]

Miscellaneous ExerciseArea under simple curves (vertical strips) · Miscellaneous Exercise
The question

1. Find the area under the given curves and given lines:

(i) y=x2y = x^2⁠, x=1x = 1⁠, x=2x = 2 and xx-axis

(ii) y=x4y = x^4⁠, x=1x = 1⁠, x=5x = 5 and xx-axis

Exam answer

(i) (figure) The region is bounded by y=x2y = x^2⁠, the xx-axis and the lines x=1x = 1⁠, x=2x = 2⁠.

The lines meet the curve at (1,1)(1, 1) and (2,4)(2, 4)⁠. On [1,2][1, 2]⁠, y=x2>0y = x^2 > 0⁠.

A1=∫12y dx=∫12x2 dx=[x33]12=83−13\begin{aligned} A_1 &= \int_1^2 y\,dx = \int_1^2 x^2\,dx \\ &= \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} \end{aligned}

∴\therefore Required area =73= \frac{7}{3} sq. units.

(ii) (figure) The region is bounded by y=x4y = x^4⁠, the xx-axis and the lines x=1x = 1⁠, x=5x = 5⁠.

The lines meet the curve at (1,1)(1, 1) and (5,625)(5, 625)⁠. On [1,5][1, 5]⁠, y=x4>0y = x^4 > 0⁠.

A2=∫15y dx=∫15x4 dx=[x55]15=31255−15\begin{aligned} A_2 &= \int_1^5 y\,dx = \int_1^5 x^4\,dx \\ &= \left[\frac{x^5}{5}\right]_1^5 = \frac{3125}{5} - \frac{1}{5} \end{aligned}

∴\therefore Required area =31245=624.8= \dfrac{3124}{5} = 624.8 sq. units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Cut the region into thin vertical strips. A strip at xx has height yy and width dxdx⁠, so its area is y dxy\,dx⁠, and the integral ∫aby dx\int_a^b y\,dx adds up all the strips from x=ax = a to x=bx = b⁠. Both curves here stay above the xx-axis on the given interval, so each area is one integral with no splitting and no bars: (i) ∫12x2 dx=73\int_1^2 x^2\,dx = \frac73 and (ii) ∫15x4 dx=31245=624.8\int_1^5 x^4\,dx = \frac{3124}{5} = 624.8⁠.

Given, to find, and the plan
Given
(i) The curve y=x2y = x^2⁠, the lines x=1x = 1 and x=2x = 2⁠, and the xx-axis. (ii) The curve y=x4y = x^4⁠, the lines x=1x = 1 and x=5x = 5⁠, and the xx-axis.
To find
The area of each region, in square units.
Plan
For each part: (1) sketch the region and find where the lines meet the curve; (2) check that the curve is above the xx-axis on the interval; (3) write the area as ∫aby dx\int_a^b y\,dx⁠; (4) integrate with the power rule and put in the limits. In (ii), also show that the exact value 31245\frac{3124}{5} is the key's decimal 624.8.

Basics you need

1Area as a sum of thin strips

For a curve y=f(x)y = f(x) above the xx-axis, the area between the curve, the xx-axis and the lines x=ax = a⁠, x=bx = b is A=∫aby dx=∫abf(x) dxA = \int_a^b y\,dx = \int_a^b f(x)\,dx (Section 8.2). Basics: a strip has area y dx

2A definite integral is a number

If F′(x)=f(x)F'(x) = f(x)⁠, then ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)⁠, written [F(x)]ab\big[F(x)\big]_a^b⁠: upper limit first, then subtract the lower. Basics: the definite integral

3The power rule

∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C for n≠−1n \ne -1 (Chapter 7, Section 7.2). So ∫x2 dx=x33+C\int x^2\,dx = \frac{x^3}{3} + C and ∫x4 dx=x55+C\int x^4\,dx = \frac{x^5}{5} + C⁠. Lesson: area under a curve

4Why we check the sign first

Where a curve is below the xx-axis, y dxy\,dx is negative, and that part must be taken as a size. Here we check that this never happens. Basics: area below the axis

Part (i): y=x2y = x^2 from x=1x = 1 to x=2x = 2

  1. Sketch the region and find the corner points

    y=x2y = x^2 is the parabola with its lowest point at the origin, opening upwards. The line x=1x = 1 meets it where y=12=1y = 1^2 = 1⁠, at (1,1)(1, 1)⁠; the line x=2x = 2 meets it where y=22=4y = 2^2 = 4⁠, at (2,4)(2, 4)⁠. The region is the shaded part below the curve, above the xx-axis, between these two lines.

    x y 1 2 1 4 O (1, 1) (2, 4) 7/3 y = x² x = 1 x = 2
    The region under y=x2y = x^2 from x=1x = 1 to x=2x = 2 (blue). Its corners on the curve are (1,1)(1, 1) and (2,4)(2, 4)⁠. The two axes have different scales.
    The sketch shows which boundary is on top (the curve), which is at the bottom (the xx-axis) and where the region starts and stops (x=1x = 1 and x=2x = 2⁠). In the board exam the figure itself earns a mark.
    Lesson: the area under a curve

  2. Check that the curve is above the xx-axis

    For 1≤x≤21 \le x \le 2⁠, x2≥1>0x^2 \ge 1 > 0⁠. So the whole region lies above the xx-axis, and no part of the integral can come out negative.

    Only a part of the curve below the axis needs bars (its integral is negative). There is none here, so a single integral gives the area.
    Basics: area below the axis

  3. Write the area as an integral

    A1=∫12y dx=∫12x2 dx.A_1 = \int_1^2 y\,dx = \int_1^2 x^2\,dx .
    A thin vertical strip at xx has height y=x2y = x^2 (from the axis up to the curve) and width dxdx⁠. Adding the strips from x=1x = 1 to x=2x = 2 is the definite integral.
    Basics: a strip has area y dx

  4. Integrate and put in the limits

    A1=[x33]12=233−133=83−13=73.\begin{aligned} A_1 &= \left[\frac{x^3}{3}\right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} \\ &= \frac{8}{3} - \frac{1}{3} = \frac{7}{3} . \end{aligned}

    So the area is 73\frac{7}{3} sq. units, about 2.3333 sq. units.

    By the power rule with n=2n = 2⁠, an antiderivative of x2x^2 is x33\frac{x^3}{3} (its derivative is 3x23=x2\frac{3x^2}{3} = x^2⁠). The fundamental theorem says the integral is its value at the upper limit minus its value at the lower limit. No "+ C" is needed: it would cancel in the subtraction.
    Basics: the definite integral is F(b) − F(a)

Part (ii): y=x4y = x^4 from x=1x = 1 to x=5x = 5

  1. Sketch the region and find the corner points

    y=x4y = x^4 is flat near the origin and then rises very steeply. The line x=1x = 1 meets it at (1,1)(1, 1)⁠, since 14=11^4 = 1⁠. The line x=5x = 5 meets it at (5,625)(5, 625)⁠, since 54=5⋅5⋅5⋅5=6255^4 = 5 \cdot 5 \cdot 5 \cdot 5 = 625⁠.

    x y 1 2 3 4 5 200 400 600 O (1, 1) (5, 625) 3124/5 y = x⁴
    The region under y=x4y = x^4 from x=1x = 1 to x=5x = 5 (blue). The yy-axis is squeezed so that the height 625 fits: one unit across is 85 px, while 100 units up is only 38 px. That is why the curve looks almost flat near (1,1)(1, 1)⁠.
    With equal scales the point (5,625)(5, 625) would be 125 times higher than it is wide, so the picture must use a smaller scale upwards. The shape of the region (curve on top, axis below, between x=1x = 1 and x=5x = 5⁠) is what the sketch needs to show.
    Lesson: the area under a curve

  2. Check the sign and write the area as an integral

    For 1≤x≤51 \le x \le 5⁠, x4≥1>0x^4 \ge 1 > 0⁠, so the region is above the xx-axis and

    A2=∫15y dx=∫15x4 dx.A_2 = \int_1^5 y\,dx = \int_1^5 x^4\,dx .
    As in Part (i), Steps 2 and 3: the strip at xx has height x4x^4 and width dxdx⁠, and nothing lies below the axis, so one integral is the area.
    Basics: a strip has area y dx

  3. Integrate and put in the limits

    A2=[x55]15=555−155=31255−15=31245.\begin{aligned} A_2 &= \left[\frac{x^5}{5}\right]_1^5 = \frac{5^5}{5} - \frac{1^5}{5} \\ &= \frac{3125}{5} - \frac{1}{5} = \frac{3124}{5} . \end{aligned}
    By the power rule with n=4n = 4⁠, an antiderivative of x4x^4 is x55\frac{x^5}{5} (its derivative is 5x45=x4\frac{5x^4}{5} = x^4⁠). Here 55=625⋅5=31255^5 = 625 \cdot 5 = 3125⁠. It is tempting to simplify 31255=625\frac{3125}{5} = 625 first: that also works, giving 625−15=31245625 - \frac15 = \frac{3124}{5}⁠.
    Basics: the definite integral is F(b) − F(a)

  4. The exact value and the key's decimal are the same number

    Divide: 3124=5⋅624+43124 = 5 \cdot 624 + 4⁠, so

    31245=624+45=624+810=624.8.\frac{3124}{5} = 624 + \frac{4}{5} = 624 + \frac{8}{10} = 624.8 .

    So the area is 31245=624.8\frac{3124}{5} = 624.8 sq. units exactly. The answer key's 624.8 is not a rounded value; it is the exact value written as a decimal.

    A fraction whose denominator (in lowest terms) has no prime factors other than 2 and 5 has a decimal that stops. Here the denominator is 5, and 45=810\frac45 = \frac{8}{10}⁠, so the decimal stops after one place. (Compare Part (i): 73=2.333…\frac73 = 2.333\ldots never stops, which is why the key gives it as a fraction.)
    624.8×5=3124624.8 \times 5 = 3124⁠, confirming the division.
Result

(i) The area under y=x2y = x^2 between x=1x = 1 and x=2x = 2 is 73\frac{7}{3} sq. units. (Textbook answer: 73\frac73⁠.)

(ii) The area under y=x4y = x^4 between x=1x = 1 and x=5x = 5 is 31245=624.8\frac{3124}{5} = 624.8 sq. units. (Textbook answer: 624.8, the same number, since 31245=624.8\frac{3124}{5} = 624.8 exactly.)

Check it with numbers

(i) A computer's numerical integration of x2x^2 over [1,2][1, 2] gives 2.3333333333, and 73≈2.3333333333\frac73 \approx 2.3333333333 ✓. A rough check: one strip using the height at the middle, 1.52×1=2.251.5^2 \times 1 = 2.25⁠, is a little too small, and the trapezium 1+42×1=2.5\frac{1 + 4}{2} \times 1 = 2.5 is a little too large (the curve bends upwards); 73≈2.33\frac73 \approx 2.33 lies between them ✓.

(ii) Numerical integration of x4x^4 over [1,5][1, 5] gives 624.8000000000 ✓. Four strips of width 1 using the middle heights give 1.54+2.54+3.54+4.541.5^4 + 2.5^4 + 3.5^4 + 4.5^4 =5.0625+39.0625+150.0625+410.0625= 5.0625 + 39.0625 + 150.0625 + 410.0625 =604.25= 604.25⁠, a little too small; the trapezium 1+6252×4=1252\frac{1 + 625}{2} \times 4 = 1252 is far too large. 624.8 lies between them ✓.

Common mistakes
  1. Subtracting the wrong way round: 13−83=−73\frac13 - \frac83 = -\frac73⁠. The bracket is always (value at the upper limit) minus (value at the lower limit); a negative answer for a region above the axis means the order is wrong.
  2. Putting the limits into yy instead of the antiderivative: 54−14=6245^4 - 1^4 = 624 is the difference of heights, not an area.
  3. Forgetting to divide by n+1n + 1: [x5]15=3124\big[x^5\big]_1^5 = 3124 is five times too big.
  4. Writing "≈ 624.8" or leaving out "sq. units". The value is exactly 624.8, and an area needs its units.
What to take away
  1. For a curve above the xx-axis between x=ax = a and x=bx = b⁠, the area is ∫aby dx\int_a^b y\,dx⁠.
  2. Check the sign of yy on the interval first: if it never goes below the axis, one integral is enough. Compare Miscellaneous Q4, where y=x3y = x^3 dips below.
  3. Find the corner points by putting each line x=ax = a into the curve; they belong in the sketch.

Every fact used, and where it comes from

FactUsed inSource
Strip area dA=y dxdA = y\,dx⁠; A=∫aby dxA = \int_a^b y\,dxPart (i), Step 3; Part (ii), Step 2Section 8.2 (Lesson)
A part below the xx-axis gives a negative integral and is taken as a sizePart (i), Step 2Section 8.2, Remark (Fig 8.3) (Basics)
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n≠−1n \ne -1Part (i), Step 4; Part (ii), Step 3Chapter 7, standard formula (i), Section 7.2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Part (i), Step 4; Part (ii), Step 3Chapter 7, Theorem 2, Section 7.8.3 (Basics)
Fractions with denominator 5 have terminating decimalsPart (ii), Step 4Arithmetic (decimal expansions)
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