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AI HL · AHL 3

Further Integration

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AHL 3.6

Mixed practice: definite integrals and areas

AHL 3.6

Mixed practice: definite integrals and areas

Target: I can find exact values of definite integrals of \(ce^{-kx}\) and of \(\frac{c}{ax+b}\), including with negative limits. (+5 more)

By the end of this section

Key points

Exact integrals

  • \(\int e^{kx}\,dx=\frac1ke^{kx}\). Divide by the number in front of x. Then \(e^{k\ln a}=a^k\), so a ln limit gives a power of a.
  • \(\int\frac1x\,dx=\ln|x|\). When both limits are negative, take the modulus of each limit: \(\int_{-8}^{-2}\frac1x\,dx=\ln2-\ln8\).
  • \(\int\frac1{ax+b}\,dx=\frac1a\ln|ax+b|\). A negative denominator, as in \(\frac1{x-5}\) on [2, 3], gives a negative integral: the curve is below the axis.
  • \(\int_k^{2k}\frac1x\,dx=\ln2k-\ln k=\ln2\) for \(k>0\), whatever k is.
  • \(\int x\sqrt{x^2+c}\,dx=\frac13(x^2+c)^{3/2}\): differentiate \((x^2+c)^{3/2}\) to see why. This is the reverse of the chain rule.

Areas between curves and under a graph

  • Area between two curves = integral of (upper − lower). If the curves cross, split at the crossing and use upper − lower on each part.
  • When the region sits under two curves, it is under the lower one: use one curve up to the crossing and the other after it.
  • Where a graph goes below the axis, \(\int f\) is negative. The area is \(\int|f|\). For \(\sin2x\) on \([0,\pi]\), find each lobe separately, then add the positive values.
  • On a graph: signed areas add and subtract, areas (with |f|) only add.

Properties and unknown limits

  • \(\int_a^bkf(x)\,dx=k\int_a^bf(x)\,dx\) and \(\int_a^bc\,dx=c(b-a)\).
  • \(f(2x)\) is f squeezed to half the width. If the limits are halved too, \(\int_0^{b/2}f(2x)\,dx=\frac12\int_0^bf(x)\,dx\).
  • To find an unknown limit: do the integral with the unknown in the limits, set it equal to the value, then solve. Reject any root the question rules out.
  • A "hence" question wants you to use the result you have just found.

On the calculator

Numerical integral

  • Enter the upper curve minus the lower curve as the function.
  • Type the limits. Use radians for trig graphs.
  • Give the area to 3 significant figures.

Screenshot and video to come

Intersection of two curves

  • Graph both curves in a window that shows the crossing.
  • Use the intersection tool, or solve the equation \(f(x)=g(x)\).
  • Use the x-values as the limits of the area.

Screenshot and video to come

Solving for a limit

  • Integrate by hand, with a in the limits, to get an equation in a.
  • Solve it with the polynomial solver.
  • Reject any root the question rules out, such as a negative a.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
An exponential with a log limit

Find the exact value of \(\int_0^{\ln 4} 3e^{-2x}\,dx\), then give it to 3 significant figures.

Working
  1. Integrate

    \(\int 3e^{-2x}\,dx=-\frac{3}{2}e^{-2x}\)

    Divide by the number in front of x in the power, which is −2.

  2. Limits

    \(\left[-\frac32 e^{-2x}\right]_0^{\ln 4}=-\frac32 e^{-2\ln 4}+\frac32\)

    The lower limit gives \(e^0=1\).

  3. Remove the ln

    \(e^{-2\ln 4}=4^{-2}=\frac1{16}\)

    \(e^{k\ln a}=a^k\). A power of e cancels the ln.

  4. Exact value

    \(\frac32\left(1-\frac1{16}\right)=\frac{45}{32}=1.41\)

    Keep the fraction for the exact answer.

Video walkthrough coming
1/x between negative limits

Find the exact value of \(\int_{-8}^{-2}\frac6x\,dx\), then give it to 3 significant figures.

Working
  1. Integrate

    \(\int\frac6x\,dx=6\ln|x|\)

    The modulus is needed because x is negative: ln of a negative number does not exist.

  2. Limits

    \(6\ln|-2|-6\ln|-8|=6\ln2-6\ln8\)

    Take the modulus of each limit, then upper minus lower.

  3. Combine

    \(6\ln\frac28=6\ln\frac14=-6\ln4=-12\ln2\)

    \(\ln\frac1a=-\ln a\). A negative answer is right: the curve is below the axis.

  4. Value

    \(-12\ln 2=-8.32\)

Video walkthrough coming
1/(px + q)

Find the exact value of \(\int_1^5\frac4{3x+1}\,dx\), then give it to 3 significant figures.

Working
  1. Integrate

    \(\int\frac4{3x+1}\,dx=\frac43\ln|3x+1|\)

    Divide by 3, the number in front of x.

  2. Limits

    \(\frac43\ln16-\frac43\ln4\)

    At x = 5, 3x + 1 = 16. At x = 1, 3x + 1 = 4.

  3. Combine

    \(\frac43\ln\frac{16}{4}=\frac43\ln4=1.85\)

    Exact form: \(\frac43\ln 4\), or \(\frac83\ln2\).

Video walkthrough coming
A denominator that is negative

Evaluate \(\int_1^4\frac3{x-6}\,dx\).

Working
  1. Integrate

    \(\int\frac3{x-6}\,dx=3\ln|x-6|\)

    On [1, 4] the value of x − 6 is negative. Use the modulus.

  2. Limits

    \(3\ln|-2|-3\ln|-5|=3\ln2-3\ln5\)

    x − 6 is −2 at x = 4 and −5 at x = 1.

  3. Value

    \(3\ln\frac25=-2.75\)

    Negative: the whole curve lies below the axis.

Video walkthrough coming
Solving for the limit

Given that \(a>0\) and \(\int_a^{2a}(4x-1)\,dx=22\), find a.

Working
  1. Integrate

    \(\left[2x^2-x\right]_a^{2a}\)

    Work out the integral with a in the limits first.

  2. Limits

    \((8a^2-2a)-(2a^2-a)=6a^2-a\)

    Substitute 2a, then a. Subtract.

  3. Equation

    \(6a^2-a=22\), so \(6a^2-a-22=0\)

    Now it is a quadratic equation.

  4. Solve for a

    Polynomial solver: a = 2 or a = −1.83

    Reject the negative root: a > 0. So a = 2.

Video walkthrough coming
Integral of f(2x)

Given that \(\int_0^{10}f(x)\,dx=14\), find \(\int_0^{5}4f(2x)\,dx\).

Working
  1. The inside

    \(f(2x)\) is \(f\) with x squeezed to half the width

    When x runs from 0 to 5, 2x runs from 0 to 10. That matches the given limits.

  2. The width

    \(\int_0^5f(2x)\,dx=\frac12\int_0^{10}f(x)\,dx=7\)

    The area has half the width, so it is half as big.

  3. The multiple

    \(\int_0^54f(2x)\,dx=4\times7=28\)

    A constant multiple comes outside.

Video walkthrough coming
Area split where the curves cross

The graphs of \(y=\cos 2x\) and \(y=\sin x\) are shown for \(0\le x\le\frac\pi2\). They meet where \(x=\frac\pi6\). Find the exact area of the region between the graphs for \(0\le x\le\frac\pi2\).

0.511.5−1−0.50.51xyy = cos 2xy = sin xP
Working
  1. Which is on top

    \(\cos2x\) above \(\sin x\) for \(0\le x\le\frac\pi6\); \(\sin x\) above for \(\frac\pi6\le x\le\frac\pi2\)

    Read it from the graph. The upper curve changes at P.

  2. First part

    \(\int_0^{\pi/6}(\cos2x-\sin x)\,dx=\left[\tfrac12\sin2x+\cos x\right]_0^{\pi/6}=\frac{3\sqrt3}{4}-1\)

    Upper minus lower.

  3. Second part

    \(\int_{\pi/6}^{\pi/2}(\sin x-\cos2x)\,dx=\left[-\cos x-\tfrac12\sin2x\right]_{\pi/6}^{\pi/2}=\frac{3\sqrt3}{4}\)

    Now the other curve is upper.

  4. Add

    \(\frac{3\sqrt3}{4}-1+\frac{3\sqrt3}{4}=\frac{3\sqrt3}{2}-1=1.60\)

    Both parts are areas, so add them.

Video walkthrough coming
Two trig graphs of x squared

(a) Sketch \(y=\cos(x^2)\) and \(y=\sin(2x^2)\) for \(0\le x\le1.2\). (b) The graphs and the y-axis enclose a region. Find its area.

0.20.40.60.811.2−1−0.50.51xyy = cos(x²)y = sin(2x²)P
Working
  1. Sketch

    From the calculator: graph both, radians mode, window 0 ≤ x ≤ 1.2

    Both start at x = 0: cosine at 1 and sine at 0. They cross once, at P.

  2. Crossing

    Solve \(\cos(x^2)=\sin(2x^2)\): x = 0.7236

    Use the intersection tool, or solve for x. This is the right-hand edge of the region.

  3. Area

    From the calculator: \(\int_0^{0.7236}\left(\cos(x^2)-\sin(2x^2)\right)dx=0.471\)

    Upper minus lower. Cosine is above on this interval.

Video walkthrough coming