AI HL · AHL 3
0 of 78 skills done
Area between a curve and the y-axis
AHL 3.7
Screenshot and video to come
Screenshot and video to come
The region lies between the curve and the y-axis. Drag the points c and d up and down the curve, and see the integral of x with respect to y.
The thin lines are horizontal strips. Select a point (Tab) and use the up and down arrow keys.
Drag c or d to change the region.
Find the area enclosed by the curve \(y=\ln x\), the y-axis and the lines \(y=2\) and \(y=3\).
Have a go first. Then open the working one line at a time.
Strip between y = 2 and y = 3, to the left of the curve
The area is the width of the strip, x, added up from y = 2 to y = 3.
\(y=\ln x \Rightarrow x=e^{y}\)
Undo the ln with e.
\(\text{Area}=\int_{2}^{3}e^{y}\,dy\)
The limits are the y-values given in the question.
From the calculator: \(\int_{2}^{3}e^{y}\,dy = 12.7\)
Give 3 significant figures.
Find the area between the curve \(y=\sqrt{x}\), the y-axis and the lines \(y=2\) and \(y=5\).
Have a go first. Then open the working one line at a time.
\(y=\sqrt{x}\Rightarrow x=y^{2}\)
Square both sides.
\(\text{Area}=\int_{2}^{5}y^{2}\,dy\)
Limits are y-values.
From the calculator: \(\int_{2}^{5}y^{2}\,dy = 39\)
This is also \(\frac{5^3-2^3}{3}=39\).
Find the area between the curve \(y=x^{3}\), the y-axis and the lines \(y=2\) and \(y=10\).
Have a go first. Then open the working one line at a time.
\(y=x^{3}\Rightarrow x=\sqrt[3]{y}=y^{1/3}\)
Take the cube root.
\(\text{Area}=\int_{2}^{10}y^{1/3}\,dy\)
Limits are y-values.
From the calculator: \(\int_{2}^{10}y^{1/3}\,dy = 14.3\)
3 significant figures.
Find the area between the curve \(y=e^{4x}\), the y-axis and the lines \(y=2\) and \(y=7\).
Have a go first. Then open the working one line at a time.
\(y=e^{4x}\Rightarrow \ln y=4x\Rightarrow x=\tfrac{1}{4}\ln y\)
Take ln of both sides, then divide by 4.
\(\text{Area}=\int_{2}^{7}\tfrac{1}{4}\ln y\,dy\)
Limits are y-values.
From the calculator: \(\int_{2}^{7}\tfrac{1}{4}\ln y\,dy = 1.81\)
3 significant figures.
Find the area between the curve \(y=\frac{5}{x-4}\), the y-axis and the lines \(y=3\) and \(y=7\).
Have a go first. Then open the working one line at a time.
\(y=\frac{5}{x-4}\Rightarrow x-4=\frac{5}{y}\Rightarrow x=\frac{5}{y}+4\)
Swap the roles of x and y, then solve for x.
\(\text{Area}=\int_{3}^{7}\left(\frac{5}{y}+4\right)dy\)
The strip goes from the y-axis (x = 0) to the curve, so it includes the part to the left of x = 4.
From the calculator: \(\int_{3}^{7}\left(\frac{5}{y}+4\right)dy = 20.2\)
3 significant figures.
Find the area between the curve \(y=4\sin x\), the y-axis and the lines \(y=1\) and \(y=3\), for \(0\le x\le\frac{\pi}{2}\).
Have a go first. Then open the working one line at a time.
\(\sin x=\frac{y}{4}\Rightarrow x=\sin^{-1}\!\left(\frac{y}{4}\right)\)
Divide by 4, then use the inverse sine. On \(0\le x\le\frac{\pi}{2}\) there is one such x for each y.
\(\text{Area}=\int_{1}^{3}\sin^{-1}\!\left(\frac{y}{4}\right)dy\)
Limits are y-values. Check the calculator is in radian mode.
From the calculator: \(\int_{1}^{3}\sin^{-1}\!\left(\frac{y}{4}\right)dy = 1.06\)
3 significant figures.
Let \(a>1\). (a) Find the area between the curve \(y=\ln x\), the y-axis and the lines \(y=0\) and \(y=\ln a\), in terms of a. (b) Use the rectangle with corners \((0,0)\) and \((a,\ln a)\) to show that \(\int_{1}^{a}\ln x\,dx=a(\ln a-1)+1\).
Have a go first. Then open the working one line at a time.
\(x=e^{y}\)
The curve is y = ln x.
\(\int_{0}^{\ln a}e^{y}\,dy=\left[e^{y}\right]_{0}^{\ln a}=a-1\)
e to the ln a is a, and e⁰ is 1.
\(a\ln a=\int_{1}^{a}\ln x\,dx+(a-1)\)
The rectangle is split in two: the area under the curve from x = 1 to x = a, and the area to the y-axis.
\(\int_{1}^{a}\ln x\,dx=a\ln a-a+1=a(\ln a-1)+1\)
Rearrange. This is the result to show.