Video: Finding the Area under the Curve of a Polynomial Function

Calculate the area bounded by the graph of the function 𝑓(π‘₯) = (5 βˆ’ π‘₯)(π‘₯ βˆ’ 1)Β² and the two coordinate axes.

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Video Transcript

Calculate the area bounded by the graph of the function 𝑓 of π‘₯ is equal to five minus π‘₯ multiplied by π‘₯ minus one squared and the two coordinate axes.

The question wants us to calculate the area bounded by our function 𝑓 of π‘₯ is equal to five minus π‘₯ multiplied by π‘₯ minus one squared and our two coordinate axes. So to find the area bounded by a curve and the coordinate axes, we recall that, for a function 𝑓, which is continuous on a closed interval from π‘Ž to 𝑏, and this function 𝑓 of π‘₯ is greater than or equal to zero on this closed interval. Then the area bounded by our curve 𝑦 is equal to 𝑓 of π‘₯, the lines π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏 and the π‘₯-axis is given by the integral from π‘Ž to 𝑏 of 𝑓 of π‘₯ with respect to π‘₯.

In fact, if our function 𝑓 of π‘₯ was less than or equal to zero on a closed interval from π‘Ž to 𝑏, if we multiply our function by negative one, we now have a new function which is greater than or equal to zero on the closed interval from π‘Ž to 𝑏. This means we can then calculate the area in the same way. Instead, we get a negative result which we just multiplied by negative one. So because our function 𝑓 of π‘₯ is a polynomial and it’s continuous on all of the real numbers, we can use this method to calculate the area bounded by a curve and the two coordinate axes. Let’s start by sketching a graph of our region.

We start by finding the 𝑦-intercept. That’s when π‘₯ is equal to zero. This gives us five minus zero multiplied by zero minus one squared, which is equal to five. So our function 𝑓 of π‘₯ has a 𝑦-intercept at five. Next, we want to find the π‘₯-intercepts of our function 𝑓 of π‘₯, and we see that our function is already fully factored. Solving each of our factors is equal to zero gives us that we have two π‘₯-intercepts. One when π‘₯ is equal to five and another when π‘₯ is equal to one.

In fact, since our second factor is squared, we have what’s called a repeated root. And we know when we have a repeated root, our polynomial will only touch the axis at this point. Finally, if we were to expand our parentheses and distribute the exponents, we would see that our highest power of π‘₯ is π‘₯ cubed and its coefficient is negative one. This means that our polynomial 𝑓 of π‘₯ is a cubic with a negative first term. So we’ll draw it with a similar shape. So this gives us our sketch of 𝑦 is equal to 𝑓 of π‘₯. It has a 𝑦-intercept when 𝑦 is equal to five, a repeated root when π‘₯ is equal to one, and a final π‘₯-intercept when π‘₯ is equal to five.

We want to calculate the area bounded by a curve and by the coordinate axes. And from our sketch, we can see that this region is bounded by the line π‘₯ is equal to five and the 𝑦-axis, which is the same as the line π‘₯ is equal to zero. We might be tempted to calculate this area by using two regions for our integral. However, we notice from our sketch that the function 𝑓 of π‘₯ is greater than or equal to zero on the closed interval from zero to five. Therefore, we can calculate the area of our region, which is the same as the area bounded by the curve 𝑦 is equal to 𝑓 of π‘₯, the lines π‘₯ is equal to zero and π‘₯ is equal to five, and the π‘₯-axis. As the integral from zero to five of 𝑓 of π‘₯ with respect to π‘₯. This gives us the integral from zero to five of five minus π‘₯ multiplied by π‘₯ minus one squared with respect to π‘₯.

To evaluate this integral, we’re going to distribute over our parentheses. We’ll start by distributing the exponent over our second pair of parentheses. We can distribute the exponent over our parentheses using any method we want. For example, we could use the FOIL method or binomial expansion. This gives us π‘₯ squared minus two π‘₯ plus one. We now need to multiply five minus π‘₯ by π‘₯ squared minus two π‘₯ plus one. We’ll do this term by term.

We’ll start by multiplying five by π‘₯ squared. This gives us five π‘₯ squared. Next, we’ll multiply five by negative two π‘₯, which gives us negative 10π‘₯. Finally, we’ll multiply five by one, which is equal to five. Next, we’ll multiply negative π‘₯ by π‘₯ squared. This gives us negative π‘₯ cubed. We’ll multiply negative π‘₯ by negative two π‘₯. This gives us two π‘₯ squared. Finally, we’ll multiply negative π‘₯ by one, which gives us negative π‘₯.

We can rearrange and simplify to give us the integral from zero to five of negative π‘₯ cubed plus seven π‘₯ squared minus 11π‘₯ plus five with respect to π‘₯. We can now evaluate this integral by using the power rule for integration, which tells us for constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Ž times π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝑐. We add one to the exponent and then divide by this new exponent. This gives us negative π‘₯ to the fourth power over four plus seven π‘₯ cubed over three minus 11π‘₯ squared over two plus five π‘₯. Evaluated at the limits of our integral π‘₯ is equal to zero and π‘₯ is equal to five.

Evaluating at the upper limit of our integral π‘₯ is equal to five gives us negative five to the fourth power over four plus seven times five cubed over three minus 11 times five squared over two plus five multiplied by five. And we see, at our lower limit π‘₯ is equal to zero, all of our terms have a factor of π‘₯. So it’s equal to zero. That means to find our area, we just need to evaluate this expression, which we can calculate to give us 275 divided by 12. And since this represents an area and we’re not told any units for our measurement, we represent this as square units, to tell us that this number represents an area.

Therefore, we’ve shown that the area bounded by the graph of the function 𝑓 of π‘₯ is equal to five minus π‘₯ times π‘₯ minus one squared and the two coordinate axes is 275 divided by 12 square units.

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