AI HL · AHL 3
0 of 78 skills done
Area between a curve and the x-axis
AHL 3.5
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
Screenshot and video to come
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.
The diagram shows the graph of \(y = x^2 - 6x\). The shaded region is bounded by the curve, the x-axis and the lines \(x = 1\) and \(x = 4\). Find the area of the shaded region.
Have a go first. Then open the working one line at a time.
\(x^2 - 6x = x(x-6)\), so the curve is below the axis for \(0 < x < 6\)
The shaded region is wholly below the axis, so the integral will be negative.
\(\int_1^4 (x^2-6x)\,dx = \left[\frac{x^3}{3} - 3x^2\right]_1^4\)
Antiderivative first, then the limits.
\(= \left(\frac{64}{3} - 48\right) - \left(\frac13 - 3\right) = -24\)
Negative, as expected.
Area = \(|-24| = 24\)
An area cannot be negative.
The diagram shows the graph of \(y = x^3 - 7x^2 + 10x\), which crosses the x-axis at \(x = 0\), \(x = 2\) and \(x = 5\). Find the total area of the shaded region between \(x = 0\) and \(x = 5\).
Have a go first. Then open the working one line at a time.
The curve crosses the axis at \(x = 2\), so split the integral there
A plain integral from 0 to 5 would let the positive and negative parts cancel.
\(\int_0^2 y\,dx = \left[\frac{x^4}{4} - \frac{7x^3}{3} + 5x^2\right]_0^2 = \frac{16}{3}\)
Positive: this region is above the axis.
\(\int_2^5 y\,dx = - \frac{63}{4}\)
Negative: this region is below the axis.
Area = \(\frac{16}{3} + |- \frac{63}{4}| = \frac{253}{12}\)
Take each part positive, then add. About 21.1.
The diagram shows the graph of \(y = x^4 - 5x^2 + 4\) between \(x = -2\) and \(x = 2\). Use your calculator to find the total area of the shaded regions.
Have a go first. Then open the working one line at a time.
Area = \(\int_{-2}^{2}\left|x^4 - 5x^2 + 4\right|dx\)
The modulus turns the parts below the axis into positive areas. No need to split.
From the calculator: \(\int_{-2}^{2}\text{abs}(x^4 - 5x^2 + 4)\,dx = 8.0000\)
Enter the function inside abs( ).
Area = 8.00
3 significant figures.
The diagram shows the graph of \(y = \cos x - \sin x\) for \(0 \le x \le \pi\). Use your calculator, in radians, to find the total area between the curve and the x-axis.
Have a go first. Then open the working one line at a time.
Set the calculator to radians
The limits are in radians.
Area = \(\int_0^{\pi}\left|\cos x - \sin x\right|dx\)
The curve crosses the axis at \(x = \frac{\pi}{4}\); the modulus deals with that.
From the calculator: \(2.8284\)
Enter abs(cos(x) − sin(x)).
Area = 2.83
3 significant figures.
Evaluate \(\int_0^4 \sqrt{2x + 1}\,dx\).
Have a go first. Then open the working one line at a time.
Let \(u = 2x+1\). Then \(\frac{du}{dx} = 2\), so \(dx = \frac{du}{2}\)
The inner function is linear, so the substitution is simple.
Limits: \(x = 0\) gives \(u = 1\); \(x = 4\) gives \(u = 9\)
Change the limits and you never go back to x.
\(\int_{0}^{4}\sqrt{2x+1}\,dx = \frac{1}{2}\int_{1}^{9}u^{1/2}\,du = \frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_{1}^{9}\)
Integrate \(u^{1/2}\) as usual.
\(= \frac{1}{2}\left(\frac23(9)^{3/2} - \frac23(1)^{3/2}\right) = \frac{1}{2}\times\frac23\left(27 - 1\right)\) \(= \frac{26}{3}\)
About 8.67.
The curve \(y = 5\sin 3x\) crosses the x-axis at \(x = 0\) and \(x = \frac{\pi}{3}\). Find the exact area enclosed by this part of the curve and the x-axis.
Have a go first. Then open the working one line at a time.
\(y = 0\) when \(\sin 3x = 0\), so the lobe runs from \(x = 0\) to \(x = \frac{\pi}{3}\)
The lobe lies between two consecutive roots.
Area = \(\int_0^{\pi/3} 5\sin 3x\,dx = \left[-\frac{5}{3}\cos 3x\right]_0^{\pi/3}\)
Divide by 3 when you integrate sin 3x.
\(= -\frac{5}{3}\cos\pi - \left(-\frac{5}{3}\cos 0\right) = \frac{5}{3} + \frac{5}{3}\)
cos π = −1 and cos 0 = 1.
\(= \frac{10}{3}\) (3.33 to 3 s.f.)
Exact answer 10/3.
Find the value of a, where \(a > 4\), such that \(\int_4^a \frac{2}{\sqrt{x}}\,dx = 8\).
Have a go first. Then open the working one line at a time.
\(\int_{4}^{a} \frac{2}{\sqrt x}\,dx = \int_{4}^{a} 2x^{-1/2}\,dx = \left[4\sqrt x\right]_{4}^{a} = 4\sqrt a - 8\)
Write 1/√x as a power of x, then integrate. Keep the lower limit.
\(4\sqrt a - 8 = 8\), so \(4\sqrt a = 16\)
The integral equals 8, so solve for a.
\(\sqrt a = 4\), so \(a = 16\)
Square to find a.
The depth of a stream, \(d\) metres, at a distance \(x\) metres from its centre is modelled by \(d=1.2e^{-x^2/8}\) for \(x\ge 0\). The cross-sectional area of water from the centre to the point \(x=k\) is 2.5 m². Find k, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(\int_0^{k}1.2e^{-x^2/8}\,dx=2.5\)
The area from 0 to k is a function of k. Set it equal to 2.5.
\(2.5\div1.2=2.08\)
A rectangle of height 1.2 would need width 2.08. The curve is lower than 1.2, so k is bigger than 2.08.
From the calculator: solve \(\int_0^{k}1.2e^{-x^2/8}\,dx=2.5\) for k: \(k=2.75\)
There is no antiderivative to write, so the calculator solves the equation.
k = 2.75 m
It is bigger than 2.08, as the check said.