Menu
Sign up Sign in

AI HL · 13.5

Integration

0 of 42 skills done

0
13.5.4

Definite integrals and area under a curve

13.5.4

Definite integrals and area under a curve

Target: I can say that a definite integral has limits and gives a number, with no x and no c. (+5 more)

By the end of this section

Key points

Definite integrals

  • A definite integral has a lower limit and an upper limit: \(\int_a^b f(x)\,dx\). The answer is a number. It has no x and no c.
  • You only need to evaluate definite integrals on the calculator. Enter the whole expression, with brackets, and both limits.
  • Enter a negative power as it is, for example \(x^{-2}\). Enter a power of a bracket as it is: do not expand it first or change it.
  • The calculator also integrates \(e^x\), \(\frac{1}{x}\) and \(\sqrt{x}\). You do not need the antiderivative: write the integral, then read the value.

Area under a curve

  • When \(f(x)\ge0\), the area between the curve and the x-axis from \(x=a\) to \(x=b\) is \(A=\int_a^b y\,dx\).
  • Write the correct definite integral before you use the calculator. The integral is the working that gets the marks.
  • The limits are the x-values at the left and right edges of the shaded region. If a limit is where the curve meets the x-axis, find it first.
  • Give the answer to 3 significant figures, with units if the question has them.
  • A region can be a rectangle with the area under a curve taken away. A solid with the same cross-section all the way along has volume = cross-sectional area × length.

On the calculator

Evaluating a definite integral

  • Open the integral template and type the expression, the variable and the two limits.
  • Use brackets round the whole expression, and round any power or fraction.
  • Read the number. Write down the integral you entered, then the value.

Screenshot and video to come

Checking an area with a graph

  • Graph the curve and check the shaded region lies above the x-axis between your limits.
  • The area you find should be about the size of the shaded region: a quick check for a wrong limit.

Screenshot and video to come

Integrating an exponential or a reciprocal

  • Type the exponential with its brackets, for example \(8e^{-0.5x}\): the minus sign belongs in the power.
  • Type a reciprocal as a fraction, \(6/x\), with the limits both on the same side of 0.
  • Write down the integral you entered, then the value.

Screenshot and video to come

Explore: the integral and the area

Choose a curve. Drag a and b along the x-axis. Part of the region is below the axis: see what the integral does with it, and what the area does.

Orange shading is above the x-axis, teal shading is below it. Select a point (Tab) and use the arrow keys. For sin x the calculator is in radians.

Drag a or b to change the region.

  1. Choose x³ − 4x and set a = −2 and b = 2. What is the integral? Is the area really 0? Why do the two answers differ?
  2. Move b so the whole region is above the axis. What happens to the integral and the area?
  3. Move a and b so the whole region is below the axis. What sign has the integral? What would you write for the area?
  4. Why is it not enough to find the integral when a question asks for the area between the curve and the x-axis?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A definite integral of powers

Evaluate \(\int_{1}^{5} (2x^3-3x)\,dx\).

Working
  1. Write it down

    \(\int_{1}^{5} (2x^3-3x)\,dx\)

    The limits are 1 (bottom) and 5 (top). The answer is a number: no x and no c.

  2. Calculator

    From the calculator: \(\int_{1}^{5} (2x^3-3x)\,dx\) = 276

    Type the expression and the two limits into the integral template. Use brackets round the whole expression.

Video walkthrough coming
A negative power

Evaluate \(\int_{1}^{3} (6x^{-2}+x)\,dx\).

Working
  1. Write it down

    \(\int_{1}^{3} (6x^{-2}+x)\,dx\)

    The negative power stays as it is. Enter it as 6x^(−2) with brackets round the power.

  2. Calculator

    From the calculator: \(\int_{1}^{3} (6x^{-2}+x)\,dx\) = 8

    The calculator copes with negative powers. You do not need to find the antiderivative.

Video walkthrough coming
A power of a bracket

Evaluate \(\int_{1}^{4} (x-2)^3\,dx\).

Working
  1. Keep the bracket

    \(\int_{1}^{4} (x-2)^3\,dx\)

    Do not expand and do not change the bracket. Enter (x−2)^3 as it stands.

  2. Calculator

    From the calculator: \(\int_{1}^{4} (x-2)^3\,dx\) = 3.75

    Part of the curve lies below the x-axis here, so the integral is not an area. It is just a number.

Video walkthrough coming
Area under a quadratic

The diagram shows the curve \(y=2+3x-x^2\). Find the shaded area, between \(x=0\) and \(x=3\).

−11234−1123456xyy = 2 + 3x − x²
Working
  1. The curve is above the axis

    For \(0\le x\le3\), \(y\ge 0\)

    So the area is the definite integral of y.

  2. Write the integral first

    Area = \(\int_{0}^{3} (2+3x-x^2)\,dx\)

    Limits are the x-values at the left and right edges of the shaded region.

  3. Calculator

    From the calculator: \(\int_{0}^{3} (2+3x-x^2)\,dx\) = 10.5

    Give the area with units if the question has them.

Video walkthrough coming
Area under a curve with a negative power

The diagram shows the curve \(y=x+\dfrac{2}{x^2}\). Find the area of the shaded region between \(x=1\) and \(x=3\).

0.511.522.533.54123456xyy = x + 2/x²
Working
  1. Write the integral first

    Area = \(\int_{1}^{3} (x+2x^{-2})\,dx\)

    Write \(\frac{2}{x^2}\) as \(2x^{-2}\), or enter it as a fraction. Both give the same value.

  2. Calculator

    From the calculator: \(\int_{1}^{3} (x+2x^{-2})\,dx\) = 5.33

    The curve is above the x-axis, so the integral is the area.

Video walkthrough coming
Area under an exponential curve

The diagram shows the curve \(y=8e^{-0.5x}\). Find the shaded area, between \(x=0\) and \(x=4\).

−1123452468xyy = 8e^(−0.5x)
Working
  1. The curve is above the axis

    For \(0\le x\le4\), \(8e^{-0.5x}>0\)

    An exponential is never below the x-axis, so the area is the definite integral.

  2. Write the integral first

    Area = \(\int_{0}^{4} 8e^{-0.5x}\,dx\)

    Limits are the x-values at the left and right edges of the shaded region.

  3. Calculator

    From the calculator: \(\int_{0}^{4} 8e^{-0.5x}\,dx\) = 13.8

    You do not need the antiderivative. Put the minus sign inside the power.

Video walkthrough coming
A region above a curve, then a volume

The end face of a concrete block is the shaded region in the diagram. It lies inside the rectangle with corners \((0,0)\), \((5,0)\), \((5,10)\) and \((0,10)\), above the curve \(y=0.4x^2\). Lengths are in metres. The block is 6 m long and has the same cross-section all the way along. Find its volume.

−1123456724681012xyy = 0.4x²y = 10
Working
  1. The whole rectangle

    Rectangle: \(5\times10=50\) m²

    The shaded region is what is left of the rectangle.

  2. Area under the curve

    From the calculator: \(\int_{0}^{5} 0.4x^2\,dx\) = 16.67 m²

    This is the part of the rectangle below the curve.

  3. Shaded area

    Area = \(50-16.67\) = 33.33 m²

    Keep the unrounded value for the next step.

  4. Volume

    Volume = area × length = \(33.33\times6\) = 200 m³

    The cross-section is the same all along, so volume is the area of the end times the length.

Video walkthrough coming