Video: APCALC02AB-P1A-Q18-913101312010

What is the area of the region bounded by the curves of 𝑦 = 2π‘₯Β² and 𝑦 = βˆ’4π‘₯?

03:31

Video Transcript

What is the area of the region bounded by the curves of 𝑦 equals two π‘₯ squared and 𝑦 equals negative four π‘₯?

To find the area of a region bounded by a single or multiple curves, we use integration. If we’d just looking to find the area bounded by the single curve 𝑦 equals two π‘₯ squared between two values π‘Ž and 𝑏, then we would evaluate the definite integral of two π‘₯ squared with respect to π‘₯ with limits of π‘Ž and 𝑏. However, in this question, we’re looking to find the area bounded by two curves or a curve and a straight line, the curve 𝑦 equals two π‘₯ squared and the line 𝑦 equals negative four π‘₯ which look like this in relation to each other. So it’s the area shaded in orange that we’re looking to calculate.

We note from our sketch that the line 𝑦 equals negative four π‘₯ is above the curve in this region. So what we’ll do is find the area below the line 𝑦 equals negative four π‘₯ and then subtract the area below the curve 𝑦 equals two π‘₯ squared. But what are the limits that we’ll use for our integral? Well, in order to find these values, we first need to determine the π‘₯-coordinates of the points of intersection of these two lines. We first set the two expressions for 𝑦 equal to one another, giving two π‘₯ squared equals negative four π‘₯. We can add four π‘₯ to each side of the equation and then factor by two π‘₯ to give two π‘₯ multiplied by π‘₯ plus two is equal to zero.

We solve this equation by setting each factor in turn equal to zero, giving two π‘₯ equals zero from which we obtain π‘₯ equals zero and π‘₯ plus two equals zero from which we obtain π‘₯ equals negative two. We find then that the π‘₯-coordinates of these two points of intersection are negative two and zero. And therefore, the lower limit for our integral is negative two and the upper limit is zero. Before we actually perform the integration, we can use properties of integration to combine the two integrals together. We have that the area bounded by these two curves is given by the integral from negative two to zero of negative four π‘₯ minus two π‘₯ squared with respect to π‘₯.

Now, we’re ready to perform this integration. Remember that when we integrate powers of π‘₯, where the power is not equal to negative one, we increase the power by one and then divide by the new power. So our integral is negative four π‘₯ squared over two minus two π‘₯ cubed over three evaluated between negative two and zero. Remember there’s no need for constant of integration because this is a definite integral. And we can also cancel a factor of two in our first term. We then substitute in our limits. And we see straightaway that this integral evaluated at zero will be equal to zero.

In the second set of parentheses, negative two squared is four and multiplying by negative two gives negative eight. And the negative two cubed is negative eight multiplied by negative two gives positive 16. So we have zero minus negative eight plus 16 over three. Negative eight can be written as negative 24 over three. So the value inside the parentheses is negative eight over three. Zero minus negative eight over three is eight over three.

So we conclude that the area bounded by these two curves found by integrating the difference in their equations between the π‘₯-coordinates of their points of intersection is eight over three square units.

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