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

Calculate the area of the plane region bounded by the curve 𝑦 = 𝑥² + 6𝑥 − 7 and the 𝑥-axis.

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### Video Transcript

Calculate the area of the plane region bounded by the curve 𝑦 equals 𝑥 squared plus six 𝑥 minus seven and the 𝑥-axis.

Well, first of all, we can see the we are actually looking for the region that’s bounded by the curve and the 𝑥-axis. Well, we know that actually at the 𝑥-axis 𝑦 is going to be equal to zero. So therefore, the first thing we want to do is actually find the roots of a function. So therefore, in order to do this, what we can actually do is set our function equal to zero. So we have 𝑥 squared plus six 𝑥 minus seven equals zero.

So then, next to find our roots, what we do is we actually factor our quadratic. So we get 𝑥 plus seven multiplied by 𝑥 minus one. So just to remind us how we actually found that, we got two numbers that are actually multiplied together to give us negative seven. So positive seven and negative one give us negative seven. And then two numbers need to add together to give us positive six. So positive seven plus negative one gives us positive six.

Okay, so now let’s actually find our roots. So therefore, we can say that 𝑥 is equal to negative seven or one. And we actually found this because to find out what our roots would be we just set each of our parentheses equal to zero. So if our first parenthesis was negative seven plus seven, this gives us zero. So zero multiplied by anything gives us zero and for our second parenthesis, one minus one gives us zero, so again giving us a zero for the function.

Okay, so now we’ve got our roots. What do we do next? So now, so what this is [inaudible] to do is actually now draw a sketch of our graph. So we’ve actually got a sketch of the curve 𝑦 equals 𝑥 squared plus six 𝑥 minus seven 𝑦. And you can see the roots on the 𝑥-axis. And we know that the shape is a U-shaped parabola because actually if you look back, we’ve got a positive 𝑥 squared term. So therefore, we know it will be U. If it was a negative 𝑥 squared, it would be an inverted U.

Okay, so now, let’s look for which region now we’re trying to find. Well, we’re looking for this region here because we’re looking for the region that is actually bounded by the curve 𝑦 equals 𝑥 squared plus six 𝑥 minus seven and the 𝑥-axis itself. But how do we actually find the area of this region? Well, we actually find the area of this region using a definite integral. So we can say that the definite integral of our function with the limits 𝑏 and 𝑎.

But let’s consider why that is. So actually, if we have a small section here that I’ve marked in blue, so we can say the width of our rectangle — so the small section forms a rectangle — would be 𝑑𝑥. So it’s very small proportion of 𝑥. And our height to that point will just be equal to our function — so 𝑓 𝑥. So therefore, if we want to find the area of our rectangle that we’ve drawn, then what we do is actually multiply our two sides together — so 𝑓 𝑥 our function multiplied by 𝑑𝑥. Well, this would just give us the area of one little rectangle.

However, if we want to find the total area of our region, what we’d need is lots of these little rectangles. And what definite integral does it is that actually gives us the area of an infinite number of these rectangles. We say an infinite number because if it was any other finite number, then therefore we’d just be estimating. But actually, it gives us the complete area of this region.

Okay, so now we know what we need to do. Let’s use it to find the area of the region. So we now know that the area of the region is gonna be equal to the definite integral of 𝑥 squared plus six 𝑥 minus seven between the limits one and negative seven. Okay, but how do we actually calculate a definite integral.

So to find a definite integral, what we can say is that if we’re looking for the definite integral of a function between limits 𝑏 and 𝑎 then this is equal to the integral of our function with 𝑏 substituted in for 𝑥 minus the integral of our function with 𝑎 — so our lower limit substituted in for 𝑥. Okay, so let’s use this to find the value of our definite integral.

So the first stage is to actually integrate our function. So we get 𝑥 cubed over three plus six 𝑥 squared over two minus seven 𝑥. And just to remind us how we actually got that, we’re going to look at the second term. So if we were going to integrate six 𝑥, then what we get is six 𝑥 and then to the power of one plus one cause you actually add one to the exponent. So that gives us six 𝑥 squared. And then, you divide by the new exponent — so you divide by one plus one, so you divide by two, which gives us six 𝑥 squared over two.

So then, we actually simplify and we have 𝑥 cubed over three plus three 𝑥 squared minus seven 𝑥 and then that’s between the limits one and negative seven. Okay, so now let’s move on to the next stage.

So now, the next stage is to actually substitute in our limits for 𝑥. So we’re going to get one cubed over three plus one multiplied by one squared minus seven multiplied by one and then this is minus negative seven all cubed over three plus three multiplied by negative seven squared minus seven multiplied by negative seven which is going to be equal to negative three and two-thirds minus negative 343 over three. So be careful here, a common mistake if we got negative seven cubed, it still remains as a negative value which is negative 343 plus 147 plus 49 that’s because we had negative seven multiplied by negative seven which is a positive, which is all equal to negative 256 over three.

But it’s negative, is this right? Well, actually, we can see that the region we’re looking at is actually below the 𝑥-axis. So yes, we would expect it to give us a negative value because as I said it’s below the 𝑥-axis. So therefore, we can say that the area of our region bounded by the curve 𝑦 equals 𝑥 squared plus six 𝑥 minus seven and the 𝑥-axis is equal to 256 over three area units. And we’ve taken away the negative cause actually we’re just looking at the area of that shape. So it’s the area is 256 over three.