# Question Video: Finding the Area under the Curve of a Root Function Mathematics • Higher Education

Let π : π(π₯) = β(5π₯ + 15). Determine, to the nearest thousandth, the area bounded by the curve π¦ = π(π₯), the π₯-axis, and the line π₯ = β2.

05:31

### Video Transcript

Let π be defined by the function π of π₯ is equal to the cube root of five π₯ plus 15. Determine to the nearest thousandth the area bounded by the curve π¦ equals π of π₯, the π₯-axis, and the line π₯ equals negative two.

Now, before we go any further here, it might be sensible to sketch the curve given by π¦ equals π of π₯. First, we can find the approximate location of the π¦-intercept by setting π₯ equal to zero. When we do, we get π of zero is equal to the cube root of 15. And thatβs approximately equal to 2.5. Similarly, when we let π of π₯ equal to zero, we can find the location of any π₯-intercepts. That gives us the cube root of five π₯ plus 15 equals zero. So, we can then cube each side and subtract 15 to get five π₯ equals negative 15. And dividing through by five, we get π₯ equals negative three.

The general shape of the curve looks a lot like the cubic graph, except itβs reflected in the line π¦ equals π₯. And so, we can see that the function π of π₯ is equal to the cube root of five π₯ plus 15 looks like this. The question wants us to determine to the nearest thousandth the area bounded by our curve, the π₯-axis, and the line π₯ equals negative two. So, we add the line π₯ equals negative two to our diagram. Itβs a vertical line that passes through the π₯-axis at negative two. And so, it follows weβre going to calculate the shaded region shown.

So, how do we find the area between a curve and the π₯-axis? Well, assuming that the area lies solely above the π₯-axis and for limits π and π where π is greater than or equal to π, the area is given by the definite integral between π and π of π of π₯ with respect to π₯. Remember, if part of the area lies below the π₯-axis, then we need to be a little bit careful. And we need to split the region up and find the absolute value of the integral where it lies below the π₯-axis. In this case, though, our region lies solely above the π₯-axis. So, we can simply find the relevant integral.

This means that the area is the definite integral between negative three and negative two of the cube root of five π₯ plus 15 with respect to π₯. Now the reason weβre choosing negative three and negative two for our limits is because these are the upper and lower π₯-values where our region is to find. So, how are we going to calculate this integral? We might notice that our integrand is a composite function but that the derivative of the inner function five π₯ plus 15 is just a constant. This is a good indication to us that we can actually use integration by substitution.

Specifically, weβll let the inner function be π’. So, π’ is five π₯ plus 15. Then dπ’ by dπ₯, the derivative of this function with respect to π₯, is just five. And with integration by substitution, whilst dπ’ by dπ₯ isnβt a fraction, we do treat it a little like one. And this means this is equivalent to saying a fifth dπ’ equals dπ₯. We then replace five π₯ plus 15 with π’. And since this is the cube root, weβre going to write that as π’ to the power of one-third. Similarly, we can replace dπ₯ with a fifth dπ’. And so, our area is going to be some definite integral of a fifth π’ to the power of one-third dπ’.

But what are our upper and lower limits? To calculate these, we substitute π₯ equals negative two and π₯ equals negative three into our expression for π’. Letβs call our upper limit π’ sub one, and this is found by letting π₯ equal negative two. So, itβs five times negative two plus 15, which is equal to five. Similarly, our lower limit is found by substituting π₯ equals negative three. We get five times negative three plus 15, and thatβs equal to zero.

So, the area is going to be the definite integral between zero and five of a fifth times π’ to the power of one-third dπ’. We can take that constant factor of a fifth out if we choose, and it might make the next part a little simpler. We integrate π’ to the power of one-third with respect to π’ by adding one to the power and then dividing by that new value. So, π’ to the power of one-third becomes π’ to the power of four-thirds divided by four-thirds, which is equivalent to three-quarters π’ to the power of four-thirds.

We now need to substitute π’ equals five and π’ equals zero into this expression and find their difference. When we do, we find that our area is equal to a fifth times three-quarters times five to the power of four over three minus three-quarters times zero to the power of four over three. But of course, the second part of this difference is actually zero. So, weβre going to type into our calculator a fifth times three-quarters times five to the power of four over three. And when we do, we get 1.28248 and so on. The question, of course, asks us to determine this value to the nearest one thousandth, in other words three decimal places. Thatβs 1.282. And so, correct to the nearest thousandth, the area of the region we were looking at is 1.282 area units.