Video: Finding the Area under the Curve of a Quadratic Function

Determine the area of the plane region bounded by the curve 𝑦 = −𝑥² + 20, the 𝑥-axis, and the two lines 𝑥 = −3 and 𝑥 = 2.

04:02

Video Transcript

Determine the area of the plane region bounded by the curve 𝑦 equals negative 𝑥 squared plus 20, the 𝑥-axis, and the two lines 𝑥 equals negative three and 𝑥 equals two.

To help us actually visualize the question, what I’ve done is done a sketch here of the function 𝑦 equals negative 𝑥 squared plus 20. The first thing to note is its shape, because it’s a parabola. But the parabola is an inverted U, and this is because it’s a negative 𝑥 squared.

If it was a positive 𝑥 squared, then it’d be a U-shaped parabola. Now we actually have our curve sketched, what we want to look at is the area that we’re trying to find. So the area that we’re actually looking for is actually bounded between the lines 𝑥 equals negative three, 𝑥 equals two and the curve 𝑦 equals negative 𝑥 squared plus 20.

And in order to actually find this area, what we use is a definite integral cause that allows us to find the area beneath a curve. So what that’s allowed us to do is actually set up our definite integral because we’re gonna have the definite integral of negative 𝑥 squared plus 20 with the limits two and negative three.

And this is because these are gonna be the upper and lower limits of the region bounded by the curve and the two lines that we’re looking for. Okay, so we now know what we need to do. However, how do we actually find a definite integral?

Now in order to actually find the definite integral value, what we do is we integrate our function, which I have represented here by the capital 𝐹 for our integrated function, and then what we do is we actually substitute in our values for 𝑏 and 𝑎, so our upper and lower limits.

And when we do, we substitute them in for our 𝑥 values. And what we actually do is we get the value of our integral with the upper values, that’s 𝑏, and then we subtract from it the value of our integral with the value of 𝑎, which is our lower bound substituted in for 𝑥.

Okay, great and if we do that, we’ll get the value of our definite integral. So now what we want to do. Let’s do that, and let’s find out the value of our definite integral. So first of all, what we do is we actually integrate our function. So we get negative 𝑥 cubed over three plus 20𝑥.

And just to remind us how we did that, well we had negative 𝑥 and then we increased the exponent by one. So we went from two to three. So it’s negative 𝑥 cubed. And then we divided by the new exponent. Right! Now, the next step is to actually substitute in our upper and lower limit values for our 𝑥. So that’s our two and our negative three.

So therefore, we’re gonna get negative two cubed over three plus 20 multiplied by two and then minus. Then we’ve got negative negative three cubed over three plus 20 multiplied by negative three, which leads us to 40 minus eight over three minus negative 60 minus negative 27 over three, which gives us a 112 over three plus 51.

So therefore, we can say that the area of the plane region bounded by the curve 𝑦 equals negative 𝑥 squared plus 20, the 𝑥-axis, and the two lines 𝑥 equals negative three and 𝑥 equals two is equal to 265 over three.

And just to quickly recap what we’ve actually done to reach this answer, first of all we actually set up a definite integral. So we had the definite integral of negative 𝑥 squared plus 20 with the limits two and negative three because they are actually the bounds of the area that we were looking for.

We then actually integrated our function. And then finally, what we did is we actually substituted in our upper and lower limits for 𝑥. And then we actually subtracted the value with the lower limit in from the value with the upper limit in. And that gave us our final answer: 265 over three.

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