Question Video: Finding the Area of the Region Bounded above Two Intersected Curves and below a Straight Line

Find the area of the region bounded by the curves 𝑦 = 4 βˆ’ π‘₯Β², 𝑦 = βˆ’π‘₯, and 𝑦 = √π‘₯. Give your answer correct to one decimal place.

03:16

Video Transcript

Find the area of the region bounded by the curves 𝑦 equals four minus π‘₯ squared, 𝑦 equals negative π‘₯, and 𝑦 equals the square root of π‘₯. Give your answer correct to one decimal place.

Remember, for continuous functions 𝑓 and 𝑔, 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, as long as 𝑓 of π‘₯ is greater than or equal to 𝑔 of π‘₯ in the closed interval π‘Ž to 𝑏, is given by the integral evaluated between π‘Ž and 𝑏 of 𝑓 of π‘₯ minus 𝑔 of π‘₯. We are going to need to be a little bit careful here, as we have three curves. So, let’s begin by sketching this out and see what we’re dealing with.

The area enclosed between the three curves looks a little something like this. Now, if we’re really clever, we can actually use the definition we looked up before. We can split this region into the region above the π‘₯-axis and the region below the π‘₯-axis. We then can split this up a little bit further. We see that we have 𝑅 one, that’s the region between the π‘₯-axis and the curve 𝑦 equals root π‘₯ between π‘₯ equals zero and π‘₯ equals 𝑏. Where 𝑏 is the π‘₯-coordinate at the point of intersection of the curve 𝑦 equals root π‘₯ and 𝑦 equals four minus π‘₯ squared.

We then have 𝑅 two. That’s the region between 𝑦 equals four minus π‘₯ squared, π‘₯ equals 𝑏, and π‘₯ equals two. And the reason we’ve chosen π‘₯ equals two as our upper limit is that’s the π‘₯-value at the point at which the curve crosses the π‘₯-axis. We can even split our three up into two further regions to make life easier. But let’s deal first with the area of 𝑅 one and 𝑅 two. We need to work out the value of 𝑏.

We said it’s the π‘₯-coordinate at the point of intersection of the two curves four minus π‘₯ squared and root π‘₯. So, we set these equal to each other and solve for π‘₯. That gives us an π‘₯-value of 1.648, correct to three decimal places. So, 𝑏 is equal to 1.648. We can either do this by hand or use our graphical calculators to evaluate each of these integrals. The area of 𝑅 one becomes 1.4104 and so on. And the area of 𝑅 two is 0.23326 and so on.

Let’s now consider the area of 𝑅 three. It’s the integral between zero and two of negative π‘₯ evaluated with respect to π‘₯. We need to be a little bit careful here, since this is below the π‘₯-axis and therefore will yield a negative result on integration. In fact, it gives us negative two. So, we can say that the area is the absolute value of this. It’s two. And notice, we could have actually used the formula for area of a triangle to work this area out.

Now, the area of 𝑅 four is the definite integral between two and 𝑐 of four minus π‘₯ squared minus negative π‘₯. And here, 𝑐 is the π‘₯-coordinate of the point of intersection of the lines 𝑦 equals negative π‘₯ and 𝑦 equals four minus π‘₯ squared. We can once again set four minus π‘₯ squared equal to negative π‘₯ and solve for π‘₯. And we find, correct to three decimal places, that they intersect at the point where π‘₯ equals 2.562. We type this into our calculator and we find that the area of this region is 0.59106. We find the total of these four values, which gives us 4.2347, which is 4.2 square units, correct to one decimal place.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.