Find the area of the region under the graph of the function f on the interval [-1, 2].
f(x) = 6x + 1
bound by the x axis
Peter, if you are the "anonymous" of
http://www.jiskha.com/display.cgi?id=1291829564
you did not follow my instructions.
the line crosses at x = -1/6, so if the x-axis is the boundary, there are 2 different triangles to consider
left triangle:
base = 5/6
height = 5, Area = (1/2)(5/6)(5) = 25/12
right triangle:
base = 13/6
height = 13 , Area = (1/2)(13/6)(13) = 169/12
total = 25/12 + 169/12 = 97/6
To find the area of the region under the graph of the function f(x) = 6x + 1 on the interval [-1, 2] bounded by the x-axis, we need to set up an integral.
First, let's graph the function f(x) = 6x + 1 on the coordinate plane. This is a straight line with a slope of 6 and a y-intercept of 1.
Now, to find the area under the graph, we need to integrate the absolute value of the function from the lower limit of -1 to the upper limit of 2. Since the function is always positive in this region, we can simply integrate the function:
∫[a,b] |f(x)| dx
In this case, a = -1 and b = 2. So the integral becomes:
∫[-1, 2] |6x + 1| dx
Next, we need to break the interval [-1, 2] into two parts, because the function changes sign at x = -1. We split the integral into two parts at x = -1:
∫[-1, 0] (6x + 1) dx + ∫[0, 2] -(6x + 1) dx
Now, we can integrate each part separately:
∫[-1, 0] (6x + 1) dx = [3x^2 + x] from -1 to 0
∫[0, 2] -(6x + 1) dx = [-3x^2 - x] from 0 to 2
Evaluating these two integrals:
[3(0)^2 + 0] - [3(-1)^2 + (-1)] + [-3(2)^2 - 2] - [-3(0)^2 - 0]
Simplifying the expression:
[0 - 3(-1) - 2(2)] - [0 + 0]
= [3 + 4] - [0] = 7
So, the area of the region under the graph of f(x) = 6x + 1, bounded by the x-axis on the interval [-1, 2], is 7 square units.