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Definite Integrals and Area - Problem 2

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

The area between a curve and the x-axis is simply the definite integral. Take the definite integral from x=a to x=b by finding the integral of the function, and finding the value of the integral at b, and subtracting the value of the integral at a.

Recall that when a curve is below the x-axis, the area is negative. So, at the points where the graph is below the x-axis, you must multiply the result of the definite integral between those points by -1 so you get a positive value for the area. If necessary, split the definite integral into parts, separately integrating the function over points where it is above the x-axis and below the x-axis. It is important to remember that when taking the definite integral, you normally do not multiply part of the integral by -1. This is only done when calculating the area bounded by the graph and the x-axis.

Let’s find the area bounded by the x axis. In this example, we’re looking at the curve y equals 6 minus 3 root x and you’ll notice that, in my graph of that curve, part of the area between that curve and the x axis is above and part of it’s below. This means that we’re going to have to take this problem in parts. Because you remember that, when a curve is above the x axis, and we integrate, we get exactly the area under the curve. But when it’s below the x axis, when we integrate we get the opposite of the area.

I’m going to integrate these separately. I’m going to calculate area A1, A2 these two areas separately. So let’s do A1 first.

A1; since the curve is above the x axis here, I can just calculate these using definite integral. The integral from 0 to 4. And then my function is 6 minus root x. Actually let me write that as x to the ½, for antidifferentiation purpose. I can use the fundamental theorem of calculus to evaluate this.

Antiderivative of 6 is 6x and antiderivative for x to the ½ is x to the 3/2, divided by 3/2 which is the same as 2/3 x to the 3/2. So 3 times 2/3 is 2x to the 3/2. So this is my antiderivative for this function. I’m going to evaluate that from 0 to 4. So when I plug in 4, I get 6 times 4, 24, minus 2 times 4 to the 3/2. Is the square root of 4 which is 2, to the 3rd power which is 8 times 2, 16. So it’s 8, the area of this first region is 8.

Now the area of the second region, because this curve is below the x axis here, I’m going to have to take the opposite of the definite integral. The definite integral would give me a negative answer. So here’s what I have to do to the definite integral. I would be integrating from 4 to 9, same function. I got to put a minus sign in front. This will give me a negative answer, but putting a minus in front of it, will make that answer positive.

So minus and then, my antiderivative is again 6x minus 2x to the 3/2. I’m going to evaluate this from 4 to 9. It occurs to me that I’ve already evaluated this at 4. I plugged in 4 here and I got 8. So this is -8. I need to plug in 9 first. I plug in 9 here I get 6 times 9, 54, minus 2 times 9 to the 3/2, that’s the square root of 9 which is 3, to the 3rd power which is 27. 2 times 27, 54. So this is what I get when I plug 9 into my antiderivative.

And I already did calculate what happens when I plug in 4, I get 8. So this is going to be -8. This looks a little complicated, the minus here is from the original minus in front of the integral, and this part here, this is my antiderivative evaluated at 9. My antiderivative evaluated at 4. This is going to be 0 minus 8, -8, times -1, +8. So it appears that A2, is also 8.

Now both of these are 8. My original question was to find the area between the curve and the x axis between x equals 0 and x equals 9. So I have to add these up. My final area is A1 plus A2, which is 8 plus 8, and that’s 16.

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