When we're integrating a non-negative function from a to b where ax-axis can be thought of as the "area under the curve" of the function. However, most of the time we can't count on having a non-negative function to integrate.
Assume f is a function that's allowed to take on negative values, and we're integrating from a to b with a < b. Then 
is the weighted sum of the areas between the graph of f and the x-axis. We look at all areas between f and the x-axis. If they're on top of the x-axis we count them positively. If they're below the x-axis we count them negatively.
In other words, we add all the areas on top of the x-axis, then subtract all the areas below the x-axis.
Practice:
Let f (x) = 2x. Find  . | |
When we look at the graph, we see a big triangle below the x-axis and a little triangle above the x-axis. 
The total area above the x-axis and below f is 
The total area below the x-axis and above f is 
To get the integral of f, we take the area above the x-axis and subtract the area below the x-axis: 
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Find  | |
This graph consists of two triangles, both below the x-axis: 
The total area between the x axis and -|x| above the axis is 0. The total area between the x-axis and -|x| below the axis is 20. So 
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If  , what is a? | |
Draw a picture. We're integrating 2x from a to 1. Since the integral value is negative there must be some area below the x-axis, so we can assume a is negative. This line forms two triangles. We know the dimensions of the one above the x-axis: 
The area of the triangle above the x-axis is 1. For the triangle below the x-axis, we can say what its dimensions are in terms of a. Since a is negative, the base of the triangle is the positive distance -a. Similarly, the height of the triangle is -2a. The area of this triangle is 
We now have 
This is a perfectly reasonable equation that we can solve for a. 
Since we know a needs to be negative, a = -3 is the answer. | |
Find the integral.
where 
Answer
- This function, graphed on [-5,2], produces two rectangles:
We count the little rectangle (square) above the x-axis positively, and the big one below the x-axis negatively:
Find the integral.
Answer
- Now we have two triangles:
First we need to figure out where the line hits the x-axis, since that will tell us what the bases of the triangles are.Since
when x = 4, we know x = 4 is the point in between the two triangles.
Now we can figure out the necessary dimensions:
And find the integral:
Find the integral. (hint: draw pictures)
Hint
don't try to calculate any areas
Answer
- The hint says to not try to calculate any areas. Since the sin function is odd, the area below the x-axis and the area above the x-axis are equal:
When we count one area positively and one area negatively, the weighted areas cancel out. So
Find the integral. (hint: draw pictures)
Hint
calculate the missing area
Answer
describes half of a circle sitting on top of the x-axis. The area we want is the area inside the purple 2-by-1 rectangle, but not inside the half-circle:
The rectangle has area 2 and the half-circle has area 
, so
For a sanity check, this is approximately 0.429, which is positive, as would be expected when finding the integral of a non-negative function.
Find the integral.
where g(t) = 3t + 4.
Answer
- Here we have another triangle:
- If
and a is positive, what is a?
Answer
- This graph consists of two equally sized triangles:
The triangle on the right has area 
. Since the triangles have the same size, together they have area
So we know 
. We're given 
, so 49 and a2 must be the same. That means a = 7.
- If
and a is positive, find a.
Answer
- This is like the previous problem, only easier! From a graph we can see that the region between the graph of x and the x-axis is a triangle with area
.
So
which means
9 = a2
a = 3.
We only keep the answer a = 3 since we were told a was positive.
- Find c given that
.
Answer
- Since the integral is positive, we can assume that c is positive. If we graph the constant c on [-10,10] we get a rectangle of height c and length 20. This means the area of the rectangle is 20c. This must equal 40, which we were told was the value of the integral, so c = 2.
