Video: Finding the Area of a Region Bounded by Two Quadratic Functions

Find the area of the region bounded by the curves 𝑦 = 3π‘₯Β² βˆ’ 5π‘₯ and 𝑦 = βˆ’ 5π‘₯Β².

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Video Transcript

Find the area of the region bounded by the curves 𝑦 equals three π‘₯ squared minus five π‘₯ and 𝑦 equals negative five π‘₯ squared.

We’ll recall that the area of the region bounded by the curves 𝑦 equals 𝑓 of π‘₯, 𝑦 equals 𝑔 of π‘₯, and the lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 for continuous functions 𝑓 and 𝑔, such that 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ for all π‘₯ in the closed interval π‘Ž to 𝑏, is the definite integral between the limits of π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯. We’re, therefore, going to need to define the functions 𝑓 of π‘₯ and 𝑔 of π‘₯ really carefully and, of course, the values for π‘Ž and 𝑏, ensuring that 𝑓 of π‘₯ is greater than 𝑔 of π‘₯ in the closed interval π‘Ž to 𝑏.

The lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏 will mark the beginning and end of the region we’re interested in. So, what are the equations of these lines? They’re the π‘₯-coordinates at the points where the two graphs intercept. So, we can set the equations three π‘₯ squared minus five π‘₯ and negative five π‘₯ squared equal to each other and solve for π‘₯.

We begin by adding five π‘₯ squared to both sides. And then, we factor the expression on the left-hand side by taking out that factor of π‘₯. And we obtain π‘₯ times eight π‘₯ minus five to be equal to zero. We know that for this statement to be true, either π‘₯ itself must be equal to zero or eight π‘₯ minus five must be equal to zero. To solve this equation on the right, we add five and then divide through by eight. And we obtain π‘₯ to be equal to five-eighths.

So, we can see that the π‘₯-coordinates of the points of intersection of our two curves are zero and five-eighths. So, π‘Ž is equal to zero and 𝑏 is equal to five-eighths. Now, we’re going to need to decide which function is 𝑓 of π‘₯ and which function is 𝑔 of π‘₯. What we do next is sketch out the graphs of 𝑦 equals three π‘₯ squared minus five π‘₯ and 𝑦 equals negative five π‘₯ squared. We’re looking to establish which of the curves is essentially on top.

We know that the graph of 𝑦 equals three π‘₯ squared minus five π‘₯ is a U-shaped parabola. We can even factor the expression three π‘₯ squared minus five π‘₯, set it equal to zero, and solve for π‘₯. And we see that it passes through the π‘₯-axis at zero and five-thirds. So, it will look a little something like this. The graph of 𝑦 equals negative five π‘₯ squared is an inverted parabola which passes through the origin like this. And so, we obtain the region shaded.

We can now see that in the closed interval of zero to five-eighths, the function that’s on top, if you will, is the function defined by 𝑦 equals negative five π‘₯ squared. So, we can say that 𝑓 of π‘₯ is equal to negative five π‘₯ squared. Meaning, 𝑔 of π‘₯ is three π‘₯ squared minus five π‘₯. The area that we’re interested in must, therefore, be given by the definite integral evaluated between zero and five-eighths of negative five π‘₯ squared minus three π‘₯ squared minus five π‘₯ with respect to π‘₯.

Distributing the parentheses, and our integrand becomes negative eight π‘₯ squared plus five π‘₯. But wait a minute, we know that when we evaluate areas below the π‘₯-axis, we end up with a funny result. We get a negative value. You might wish to pause the video for a moment and consider what that means in this example. Did you work it out? We can see that our entire region sits below the π‘₯-axis and we’re just working out the difference between the areas. So, the negative results that we would obtain from integrating each function individually will simply cancel each other out. So, all that’s left is to evaluate this integral.

The integral of negative eight π‘₯ squared is negative eight π‘₯ cubed over three. And the integral of five π‘₯ is five π‘₯ squared over two. We need to evaluate this between zero and five-eighths, which is negative eight-thirds of five-eighths cubed plus five over two times five-eighths squared minus zero. That’s 125 over 384 square units. This question was fairly straightforward as the curve of 𝑦 equals negative five π‘₯ squared was greater than or equal to the curve of 𝑦 equals three π‘₯ squared minus five π‘₯ in the interval we’re interested in.

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