Video: Finding the Area of Region under a Cubic Curve

The figure shows the graph of the function 𝑓(π‘₯) = 2π‘₯Β³ βˆ’ 8π‘₯. Evaluate the area of the shaded region.

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Video Transcript

The figure shows the graph of the function 𝑓 of π‘₯ is equal to two π‘₯ cubed minus eight π‘₯. Evaluate the area of the shaded region.

We’re given a function which is a cubic polynomial. We’re also given a sketch of a region bounded by this cubic polynomial. We need to determine the area of the shaded region. The first thing we need to notice is because our function 𝑓 of π‘₯ is a cubic polynomial, this means it’s continuous. And we know how to find the area under continuous curves. We can do this by using definite integrals. However, there’s a problem. We know that using this method will give us a negative answer whenever our area lies below the π‘₯-axis, like it does in this picture.

To get around this problem, we need to split our integral into the parts where our area lies above the π‘₯-axis and where it lies below the π‘₯-axis. From the diagram, we can see that when π‘₯ is between negative two and zero, our area will be above the π‘₯-axis. And when our values of π‘₯ are between zero and two, our area will lie below the π‘₯-axis. Let’s call the area above the π‘₯-axis 𝐴 one. Because this region lies above the π‘₯-axis and between the values of π‘₯ is equal to negative two and π‘₯ is equal to zero, we know we can find this by using a definite integral. We get that 𝐴 one is equal to the definite integral from negative two to zero of two π‘₯ cubed minus eight π‘₯ with respect to π‘₯.

And we can do something similar with the area below the π‘₯-axis. Let’s call this 𝐴 two. If we tried to calculate the area of 𝐴 two using the same method, we would’ve got the integral from zero to two of two π‘₯ cubed minus eight π‘₯ with respect to π‘₯. But because this area lies below the π‘₯-axis, this will give us a negative answer. So we need to multiply this entire expression by negative one. We can do this by just writing negative one before our integral symbol. We can now evaluate both of these integrals and then add the answers together to get the total area of our shaded region.

To evaluate these integrals, we can do this term by term by using the power rule for integration. We recall this is for real constants 𝐾 and 𝑁, where 𝑁 is not equal to negative one. The integral of 𝐾 times π‘₯ to the 𝑁th power with respect to π‘₯ is equal to 𝐾 times π‘₯ to the power of 𝑁 plus one divided by 𝑁 plus one plus a constant of integration 𝐢. We add one to our exponent of π‘₯ and then divide by this new exponent. Of course, since we’re using this for a definite integral, we don’t need to add our constant of integration. It will cancel in our working out.

So let’s start by integrating the first term, two π‘₯ cubed. In this case, the exponent of π‘₯ is three. We need to add one to this exponent and then divide by this new exponent. This gives us two π‘₯ to the fourth power divided by four. And of course, we can then simplify this. Two divided by four is just equal to one-half. To integrate our second term of negative eight π‘₯, remember, we can rewrite π‘₯ as π‘₯ to the first power. Now, we need to add one to our exponent of π‘₯, which gives us two, and divide by this new exponent. This gives us negative eight π‘₯ squared divided by two.

And once again, we can simplify. Eight divided by two is equal to four. So this gave us π‘₯ to the fourth power over two minus four π‘₯ squared evaluated at the limits of integration π‘₯ is equal to zero and π‘₯ is equal to negative two. Now, we should evaluate this at the limits of integration. However, we notice if we substitute zero into this expression, both of our terms are zero. So we only need to substitute π‘₯ is equal to negative two. Remember, we need to subtract this. This gives us negative one times negative two raised to the fourth power divided by two minus four times negative two squared. This gives us negative one times eight minus 16, which we can evaluate is equal to eight.

Now we could do exactly the same thing to evaluate our second integral. However, notice we’re integrating exactly the same integrand. So when we integrate this integrand, we’ll get exactly the same expression we had before, π‘₯ to the fourth power divided by two minus four π‘₯ squared. We just need to remember we have different limits of integration and we multiplied this entire integral by negative one. So we need to evaluate negative one times π‘₯ to the fourth power divided by two minus four π‘₯ squared evaluated at the limits of integration zero and two.

Once again, when we substitute zero into both of these terms, we see we get zero. So we only need to worry about the upper limit of integration when π‘₯ is equal to two. This gives us negative one times two to the fourth power over two minus four times two squared. And if we evaluate this expression, we get eight. Finally, the area of our shaded region will be area one plus area two, which we can calculate is equal to 16. And it’s worth pointing out sometimes you’ll see this written as 16 units squared because this represents an area.

Therefore, by using integration, we were able to determine the area given to us in the graph. It was the area under the function two π‘₯ cubed minus eight π‘₯ between π‘₯ is equal to negative two and π‘₯ is equal to two. We found that this was given by 16.

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