Video: Finding the Area Between Two Curves by Integration

Find the area bounded by the curves 𝑦 = π‘₯Β² + 2 and 𝑦 = βˆ’π‘₯ + 1 between π‘₯ = βˆ’3 and π‘₯ = 4.

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Video Transcript

Find the area bounded by the curves 𝑦 equals π‘₯ squared plus two and 𝑦 equals negative π‘₯ plus one between π‘₯ equals negative three and π‘₯ equals four.

Remember, to find the area bounded by the curve 𝑦 equals 𝑓 of π‘₯, the π‘₯-axis, and the vertical lines π‘₯ equals π‘Ž and π‘₯ equals 𝑏, we evaluate the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯. Now, actually, we’re looking at the area bounded by two curves. So let’s begin by sketching this out.

The graph of 𝑦 equals π‘₯ squared plus two is a parabola that passes through the 𝑦-axis at two. And the graph of 𝑦 equals negative π‘₯ plus one is a straight line that passes through the 𝑦-axis at one. We’re looking to find the area bounded by these curves and between the vertical lines, π‘₯ equals negative three and π‘₯ equals four. So that’s the area shown. And so to find this area, we’re going to split our integrals up. We’re going to subtract the area between the π‘₯-axis and those vertical lines of the function at the bottom β€” that’s the function 𝑦 equals negative π‘₯ plus one β€” from the area between the curve of 𝑦 equals π‘₯ squared plus two, the π‘₯-axis, and those vertical lines.

And it’s important to realize that the definite integral between negative three and four of π‘₯ squared plus two with respect to π‘₯ minus the definite integral between negative three and four of negative π‘₯ plus one with respect to π‘₯. Is equal to the definite integral between negative three and four of the difference of these two functions, of π‘₯ squared plus two minus negative π‘₯ plus one with respect to π‘₯. To evaluate this, we can distribute our parentheses. And we find that our integrand becomes π‘₯ squared plus π‘₯ plus one. And then we simply integrate term by term. When we integrate π‘₯ squared, we raise the power by one and then divide by that new value. So we get π‘₯ cubed over three.

When we integrate π‘₯, similarly, we get π‘₯ squared over two. And then integrating one gives us π‘₯. We need to evaluate this between the limits of negative three and four. So that’s four cubed over three plus four squared over two plus four minus negative three cubed over three plus negative three squared over two plus negative three. That gives us an exact area of 245 over six. And if we evaluate this correct to four decimal places, we get 40.8333 or 40.8333 square units. Notice that since we rounded our answer, we’ve added a sign that means approximately equal to. The exact area is 245 over six square units, which is approximately equal to 40.8333.

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