Find the area of the region under the graph of the function f on the interval [-1, 2].
f(x) = 6x + 1
Your question is not clear at all.
When you say "under" the graph of
f(x) = 6x+1, the region would be infinitely large.
You will need to close it.
Is it bounded by the x-axis?
If so according to your interval part would be below the x-axis and part would be above it.
Make a graph to see what I mean, then clarify your question.
BTW, why would you need Calculus to find the area of a region bounded by straight lines?
To find the area of the region under the graph of a function on a given interval, you can use integration. The integral of a function represents the area between the graph and the x-axis.
First, let's find the integral of the given function f(x) = 6x + 1. To integrate, we need to find the antiderivative of the function, which is the reverse operation of differentiation.
The antiderivative of 6x is (6/2)x^2 = 3x^2, and the antiderivative of 1 is x. Therefore, the antiderivative of f(x) = 6x + 1 is F(x) = 3x^2 + x.
Now, to find the definite integral (which gives the area) of f(x) on the interval [-1, 2], we subtract the value of the antiderivative F(x) at the lower limit from the value at the upper limit: ∫[-1, 2] (6x + 1) dx = F(2) - F(-1).
Evaluating the antiderivative at these limits:
F(2) = 3(2)^2 + 2 = 12 + 2 = 14
F(-1) = 3(-1)^2 - 1 = 3 - 1 = 2
Substituting these values into the definite integral:
∫[-1, 2] (6x + 1) dx = F(2) - F(-1) = 14 - 2 = 12.
Therefore, the area of the region under the graph of the function f(x) = 6x + 1 on the interval [-1, 2] is 12 square units.