Question Video: Finding the Area of a Region Bounded by Quadratic and Linear Functions Mathematics • Higher Education

Find the area of the region bounded above by 𝑦 = 2π‘₯ and below by 𝑦 = 2π‘₯Β² βˆ’ 5π‘₯.

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Video Transcript

Find the area of the region bounded above by 𝑦 is equal to two π‘₯ and below by 𝑦 is equal to two π‘₯ squared minus five π‘₯.

In this question, we’re asked to find the area of a region bounded by the graphs of two functions, the line 𝑦 is equal to two π‘₯ and the quadratic curve 𝑦 is equal to two π‘₯ squared minus five π‘₯. And to answer questions like this, it’s always a good idea to try and sketch a graph of the area we’re asked to find. So we’ll start with our coordinate axis and we’ll sketch the line 𝑦 is equal to two π‘₯. We can see that this is given in slope–intercept form. That’s the form 𝑦 equals π‘šπ‘₯ plus 𝑏, where 𝑏 is the 𝑦-intercept and π‘š is the slope. In this case, this is a straight line passing through the origin with a slope of two. We sketch this on our diagram as shown.

Let’s now sketch the quadratic curve 𝑦 is equal to two π‘₯ squared minus five π‘₯. And there’s a few different ways we can sketch this curve. One way is to factor the equation. We can take out the shared factor of π‘₯ from the two terms to get that 𝑦 is equal to π‘₯ multiplied by two π‘₯ minus five. And we can then solve this equation equal to zero to find the coordinates of the π‘₯-intercepts. And we know for a product to be equal to zero, one of the factors must be equal to zero, so either π‘₯ is equal to zero or two π‘₯ minus five is equal to zero.

And we can solve both of these separately. We get two π‘₯-intercepts: one at π‘₯ is equal to zero and one at π‘₯ is equal to five over two. We can then mark these two π‘₯-intercepts onto our diagram. We could now determine more information about our quadratic, for example, we could find its turning point. However, it’s not necessary to answer this question. All we need to note now is that its leading coefficient is positive. And this means the shape of this curve will be a parabola which opens upwards. We get a sketch which looks like the following. And now we can see the region that we’re asked to find the area of.

There are two parts of this region. There’s the region above the π‘₯-axis below the line and above the curve, which we’ve marked in blue. And there’s also the region below the π‘₯-axis but above the curve, which we’ve marked in green. The region we’re asked to find the area of is the sum of these two areas. Since both of these functions are polynomials, they’re both continuous and we know how to integrate them. Therefore, we can find both of these areas by using integration.

In particular, we can recall the following formula for finding the area between the graphs of two functions. We know the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯ with respect to π‘₯ is the area below the curve 𝑦 is equal to 𝑓 of π‘₯ and above the curve 𝑦 is equal to 𝑔 of π‘₯ between π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏. And this formula assumes that 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ on the closed interval from π‘Ž to 𝑏. And it also assumes that we can find the areas underneath 𝑓 of π‘₯ and 𝑔 of π‘₯ between π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏.

But we know all of this information is true. For example, we can see that both of our functions are continuous for all real values, since they’re polynomials. This means that we can find their areas between these two limits by using integration. We want to use this formula to find the shaded region R. And since 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ on this interval, we can do this by using this formula. And to do this, we’re going to need to find the π‘₯-intercept of the line and the curve.

Of course, we already know the leftmost intercept. It happens when π‘₯ is equal to zero because both our line and our curve pass through the origin. To find the π‘₯-coordinate of the other intercept, we’re going to need to equate the two functions. We need to determine when two π‘₯ squared minus five π‘₯ is equal to two π‘₯. We can do this by subtracting two π‘₯ from both sides of the equation. We get two π‘₯ squared minus seven π‘₯ is equal to zero. Then we take out the shared factor of π‘₯. We get π‘₯ multiplied by two π‘₯ minus seven is equal to zero.

Then once again for a product to be equal to zero, one of the factors must be equal to zero. So either π‘₯ is equal to zero or two π‘₯ minus seven is equal to zero. And we can solve both of these for π‘₯. π‘₯ is equal to zero or π‘₯ is equal to seven over two. These give us the π‘₯-coordinates of the two intercepts between the curve and the line. π‘₯ is zero and π‘₯ is seven over two. We can add this to our diagram. And we can now use our formula to find the area of the region shaded in blue. Our lower bound will be zero and our upper bound will be seven over two. And on this entire interval, the line is above the curve. Therefore, we can find the entire shaded region by using the result, both the area in blue and the area in green.

We get the area of all of the shaded region we need to find is equal to the integral from zero to seven over two of two π‘₯ minus two π‘₯ squared minus five π‘₯ with respect to π‘₯. We can simplify our integrand by distributing the negative over the parentheses and simplifying. We get the integral from zero to seven over two of seven π‘₯ minus two π‘₯ squared with respect to π‘₯.

This is now the integral of a polynomial. And we can do this term by term by using the power rule for integration. We add one to the exponent of π‘₯ and then divide by this new exponent. This gives us seven π‘₯ squared over two minus two π‘₯ cubed over three evaluated at the limits of integration, zero and seven over two. Now all that’s left to do is evaluate our antiderivative at the limits of integration.

However, we can simplify this process by noticing if we substitute π‘₯ is equal to zero into our antiderivative, we get zero. So we only need to worry about the upper limit. Substituting this value in, we get seven times seven over two squared over two minus two multiplied by seven over two cubed over three. And if we evaluate this expression and give our answer as a mixed fraction, we get 14 and seven over 24, which is our final answer.

Therefore, we were able to show the area of the region bounded above by 𝑦 is equal to two π‘₯ and below by 𝑦 is equal to two π‘₯ squared minus five π‘₯ is 14 and seven over 24.

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