Evaluate the integral by interpreting it in terms of areas as 0 goes to 12 |2x - 7|dx?
This is my work so far:
2x-7=0
x=7/2
slope = 2; y-int = (0, -7)
Area of a triangle is 1/2BH.
I can't make sense of what you did. If you want areas, you have to do it in two integrals (triangles)
x goes from 0 to 3.5
INT (7-2x) dx
Area=1/2 b H=1/2 3.5 *7
and x goes from 3.5 to 12
int (2x-7)dx
area=1/2 b H= 1/2 (12-3.5)(24-7)
I added the two areas up and got 84.5 which is the correct. Thanks!
One quick question, for the "area=1/2 b H= 1/2 (12-3.5)(24-7)", how did you get the 24?
To evaluate the integral ∫ |2x - 7| dx from 0 to 12, we can interpret it in terms of areas.
First, let's find the x-values at which the absolute value function |2x - 7| changes sign.
|2x - 7| = 0 when 2x - 7 = 0, which gives x = 7/2.
So, we have two intervals to consider: from 0 to 7/2 and from 7/2 to 12.
In the interval [0, 7/2], the function 2x - 7 is negative. Thus, the absolute value function becomes -(2x - 7) = -2x + 7.
In the interval [7/2, 12], the function 2x - 7 is positive. So, the absolute value function remains the same, |2x - 7| = 2x - 7.
Now, let's calculate the areas of the triangles formed by the two intervals.
In the interval [0, 7/2], the triangle has a base of (7/2 - 0) = 7/2 units and a height of |-2x + 7|. The area of this triangle is (1/2) * (7/2) * |-2x + 7|.
In the interval [7/2, 12], the triangle has a base of (12 - 7/2) = 17/2 units and a height of |2x - 7|. The area of this triangle is (1/2) * (17/2) * |2x - 7|.
Now, we can set up the integral with the appropriate intervals and functions:
∫ |2x - 7| dx = ∫[0, 7/2] (-1/2)(7/2)(-2x + 7) dx + ∫[7/2, 12] (1/2)(17/2)(2x - 7) dx
Simplifying this integral will give you the final result.