Thankyou so much, I understand better now!
Would you mind showing how to solve part (b) of this question? I understand how to find the points of intersection but am slightly confused with how to find the area of the region
Sure, so the first thing to note is that our region of interest is bounded by these three curves \(y = 3x\), \(y = x^2 - 4\) and \(x = 0\). We can see that we actually have two separate areas! One is bounded above by the \(x\) axis, I labelled it as \(A_1\) and the other is bounded below by the \(x\) axis, I labelled it as \(A_2\).
Now, the bounds of \(A_1\) is from 0 to the point of intersection between \(y = 3x\) and \(y = x^2 - 4\) which you can find to be \(x = 4\). So \(A_1\) is just \(\displaystyle \int_0^4 [\text{Top } - \text{Bottom}]\,dx\), with the top curve being \(y = 3x\) and the bottom curve being \(y = x^2 - 4\). That gives us
\begin{align*}
A_1 &= \int_0^4 \left(3x - x^2 + 4\right)\,dx \\
&= \frac{3}{2}x^2 - \frac{1}{3}x^3 + 4x \bigg|_0^4 \\
&= 24 - \frac{64}{3} + 16 \\
&= 40 - \frac{64}{3}.
\end{align*}
On the other hand, \(A_2\) is a little bit harder to find. It's unclear how we could rearrange this region. But notice how this is just \(y = x^2 - 4\) bounded by the \(y\) axis! So we need to write \(x\) in terms of \(y\) which gives us \(x = \sqrt{y + 4}\). Our bounds for \(y\) go from \(y = -4\) to \(y = 0\) (hopefully you can see that on the diagram.
So \(A_2\) is just \(\int_{-4}^0 \sqrt{y + 4}\,dy\) which turns out to be \(\displaystyle \frac{16}{3}\). So the entire area is just
\begin{align*}
A &= A_1 + A_2 \\
&= 40 - \frac{64}{3} + \frac{16}{3} \\
&= 40 - \frac{48}{3} \\
&= 40 - 16 \\
&= 24\text{u}^2.
\end{align*}
If that's incorrect, pls let me know and I will try to correct it
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