The Fundamental Theorem of Calculus - Problem 1
Let’s use the fundamental theorem of calculus to solve another problem. Find the exact area bounded between y equals 6 minus x minus x² and the x axis. Here is my diagram of this region. I want to find the exact area of this region. To do so, I’m going to need to figure out exactly where the curve crosses the x axis, because these are going to be my limits of integration.
I need to set this equation equal to 0. 6 minus x minus x², that will give me the x intercepts at these points. I can factor out the -1, actually I can divide out the -1 and get x² plus x minus 6 equals 0. And then factor this. The factors will be of the form x plus or minus something, x plus or minus something. I’m seeing 2 and 3 here. S plus 3, x minus 2. -3 and 2 are my intercepts. That’s important because it tells me I have to integrate from -3 to 2.
The exact area of this region is going to be the integral from -3 to 2 of 6 minus x minus x²dx. Let’s use the fundamental theory of calculus on this. I need an antiderivative for this function. So I'll put my antiderivative in brackets here. [6x minus 1/2x² minus 1/3x³]. I need to evaluate that from -3 to 2.
First, I evaluate this thing at 2. I get 6 times 2 minus ½ times 4, minus 1/3 times 8, minus, and then I evaluate all of these at -3. 6 times -3 minus ½ times 9 minus 1/3 times -27. Here, I have 12 minus 2 is 10, minus 8/3. 10 minus 8/3, minus, here I have -18 minus 4/2. 27 divided by 3 is 9, so this is going to be +9. So -18 plus 9 is minus 9. Minus 9 minus 9/2.
Let me just simplify this a little bit more. 10 plus 9 minus 8/3 plus 9/2. So 9/2 minus 8/3. 19. I need to get a common denominator to subtract these. The common denominator will be 6, so I multiply the top and bottom by 3. This is 27/6 minus, if you multiply the top and bottom by 2, I’ll get a common denominator. So 16/6, this is 11/6. 19 plus 11/6. Which is the same as 20 and 5/6. That’s my area.