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AI HL · AHL 3

Further Integration

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AHL 3.12

Integrals under transformations of the graph

AHL 3.12

Integrals under transformations of the graph

Target: I can find the integral of kf(x), −f(x), f(x) + c and f(x) − c from the integral of f(x). (+4 more)

By the end of this section

Key points

What each transformation does to an integral

  • Suppose \(\int_a^b f(x)\,dx=X\). The new integral depends on the transformation and on the new limits.
  • Vertical stretch, scale factor k: \(kf(x)\) has integral \(kX\). With k = −1 the graph is reflected in the x-axis and the integral changes sign.
  • Horizontal stretch, scale factor s: \(f\left(\frac{x}{s}\right)\) on the interval from \(sa\) to \(sb\) has integral \(sX\). Careful: \(f(2x)\) is a stretch with scale factor \(\frac12\), so its integral over the interval from \(\frac a2\) to \(\frac b2\) is \(\frac X2\).
  • Horizontal translation: \(f(x-h)\) moves the graph h to the right, and the interval moves with it, from \(a+h\) to \(b+h\). The integral is still X.
  • Vertical translation: \(\int_a^b [f(x)+c]\,dx=X+c(b-a)\). The shift adds a strip of height c and width b − a. Write the width down first.

Combining them

  • Work in the order the question gives. Write the new limits first, then multiply the integral by each stretch factor.
  • For \(g(x)=k\,f\left(\frac{x}{s}\right)+c\) on the interval from \(sa\) to \(sb\): \(\int g\,dx=ksX+c\,s(b-a)\). The strip added by c is as wide as the new interval, not the old one.
  • A translation never changes the area. A stretch always multiplies it. Do the stretches first, then add or subtract the strip.
  • To work backwards, divide the integral of g by the product of the stretch factors.

Checking and warnings

  • Check with a simple graph: if f(x) = 1 on the interval from 0 to 2, then X = 2, and you can test any rule with it.
  • An integral is not always an area. Where the graph is below the x-axis the integral is negative.
  • The numbers in a question are not always from the same f. Use only the facts given about f.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A multiple and the negative

Given that \(\int_2^6 f(x)\,dx=7.5\), find (a) \(\int_2^6 4f(x)\,dx\); (b) \(\int_2^6 -f(x)\,dx\).

Working
  1. Take the constant out

    \(\int_2^6 4f(x)\,dx=4\int_2^6 f(x)\,dx\)

    A constant multiple comes out of the integral.

  2. Part (a)

    \(4\times 7.5=30\)

    The limits do not change, so only the factor matters.

  3. Part (b)

    \(\int_2^6 -f(x)\,dx=-7.5\)

    The factor is −1: the graph is reflected in the x-axis, so the integral changes sign.

Video walkthrough coming
Shifting the graph sideways

Given that \(\int_2^7 f(x)\,dx=18\) and \(g(x)=f(x-4)\), find \(\int_6^{11} g(x)\,dx\).

Working
  1. 246810121234567xyy = f(x)
    What g is

    \(g(x)=f(x-4)\): the graph of f moved 4 to the right.

    Every point of f moves 4 right. The interval 2 to 7 moves to 6 to 11.

  2. 246810121234567xyy = g(x)
    The new interval

    \(\int_6^{11} g(x)\,dx\)

    The question asks about 6 to 11, the image of 2 to 7.

  3. Same area

    \(\int_6^{11} g(x)\,dx=18\)

    Same shape, same area. The graphs here show one possible f; the answer does not depend on which f it is.

Video walkthrough coming
Lifting and lowering the graph

Given that \(\int_1^5 f(x)\,dx=9\), find (a) \(\int_1^5 \left[f(x)+3\right]dx\); (b) \(\int_1^5 \left[f(x)-2\right]dx\).

Working
  1. Split the integral

    \(\int_1^5 [f(x)+3]\,dx=\int_1^5 f(x)\,dx+\int_1^5 3\,dx\)

    The second part is a rectangle of height 3 and width 5 − 1 = 4.

  2. Part (a)

    \(9+3\times 4=21\)

    A vertical shift of k adds k × (b − a).

  3. Part (b)

    \(9-2\times 4=1\)

    Shifting down subtracts: k = −2.

Video walkthrough coming
Stretching along the x-axis

Given that \(\int_0^3 f(x)\,dx=4.5\): (a) the graph of g is the graph of f stretched parallel to the x-axis with scale factor 2, so \(g(x)=f\left(\frac{x}{2}\right)\). Find \(\int_0^6 g(x)\,dx\); (b) \(h(x)=f(3x)\). Find \(\int_0^1 h(x)\,dx\).

Working
  1. 12345670.511.522.53xyy = f(x)
    The graph of f

    \(\int_0^3 f(x)\,dx=4.5\)

    For example f(x) = x(3 − x) has this area.

  2. 12345670.511.522.53xyy = g(x)
    Stretch, scale factor 2

    \(\int_0^6 g(x)\,dx=2\times 4.5=9\)

    The interval 0 to 3 becomes 0 to 6: twice as wide, the same height, twice the area.

  3. Part (b)

    \(f(3x)\) is a stretch with scale factor \(\frac13\): \(\int_0^1 h(x)\,dx=\frac13\times 4.5=1.5\)

    The number inside the bracket is the reverse of the scale factor. The interval 0 to 3 becomes 0 to 1.

Video walkthrough coming
Two stretches in one

Given that \(\int_0^8 f(x)\,dx=10\), the graph of g is obtained from the graph of f by a horizontal stretch with scale factor \(\frac14\) followed by a vertical stretch with scale factor 6. Find \(\int_0^2 g(x)\,dx\).

Working
  1. Horizontal stretch

    \(8\to 8\times\tfrac14=2\), the area is multiplied by \(\tfrac14\)

    The interval 0 to 8 becomes 0 to 2. That matches the limits in the question.

  2. Vertical stretch

    The area is multiplied by 6.

    Each factor multiplies the area.

  3. Together

    \(\int_0^2 g(x)\,dx=\tfrac14\times 6\times 10=15\)

    Multiply all the factors by the original integral.

Video walkthrough coming
Stretch and shift together

Given that \(\int_0^6 f(x)\,dx=9\) and \(g(x)=2f\left(\frac{x}{3}\right)+1\), find \(\int_0^{18} g(x)\,dx\).

Working
  1. Split g

    \(\int_0^{18} g\,dx=2\int_0^{18} f\left(\tfrac{x}{3}\right)dx+\int_0^{18}1\,dx\)

    Deal with the stretched f and the constant separately.

  2. The stretched part

    \(\int_0^{18} f\left(\tfrac{x}{3}\right)dx=3\times 9=27\)

    Horizontal stretch with scale factor 3: the interval 0 to 6 becomes 0 to 18.

  3. The constant part

    \(\int_0^{18}1\,dx=18\)

    Width 18, height 1.

  4. Add

    \(2\times 27+18=72\)

    The factor 2 multiplies the stretched part only. The shift is added last.

Video walkthrough coming
Working backwards

The graph of g is obtained from the graph of f by a horizontal stretch with scale factor 4 followed by a vertical stretch with scale factor 3. Given that \(\int_0^{20} g(x)\,dx=90\), find \(\int_0^5 f(x)\,dx\).

Working
  1. Total factor

    \(4\times 3=12\)

    The integral of g is 12 times the integral of f.

  2. Write it down

    \(90=12\times\int_0^5 f(x)\,dx\)

    The interval 0 to 5 becomes 0 to 20 under the horizontal stretch, which matches.

  3. Divide

    \(\int_0^5 f(x)\,dx=90\div 12=7.5\)

    Undo the stretches by dividing.

Video walkthrough coming