The graph of a function f is shown. Let g be the function that
represents the area under the graph of f between 0 and x.
(a) Use geometry to find a formula for g(x).
(b) Verify that g is an antiderivative of f and explain how this
confirms Part 1 of the Fundamental Theorem of Calculus for
the function f
(Since I can't seem to submit a picture, the graph is f(t)=3t and I tried sketching it below)
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0 x
How would I solve this?
Oops, the graph got formatted weird after posting, sorry! The far right [ | ] would be where the x is.
if f(x) = 3t, then the area under the graph is just a triangle of base x and height 3x. So its area is g(x) = 1/2 x * 3x = 3/2 x^2
∫[0,x] 3t dt = 3/2 t^2 [0,x] = 3/2 x^2
To find the formula for g(x), we need to determine the area under the graph of f between 0 and x.
(a) The graph of f(t) = 3t is a straight line passing through the origin (0,0) with slope 3. Since it is a straight line, the area under the graph from 0 to any value x can be represented by the area of a triangle.
The formula for the area of a triangle is given by:
Area = (1/2) * base * height
In this case, the base corresponds to the value of x, and the height corresponds to the value of f(x) = 3x. Therefore, the formula for g(x) can be written as:
g(x) = (1/2) * x * f(x) = (1/2) * x * 3x = (3/2) * x^2
So, g(x) = (3/2) * x^2.
(b) To verify that g(x) is an antiderivative of f, we need to differentiate g(x) and check if the result is equal to f(x).
Differentiating g(x) with respect to x, we get:
g'(x) = (d/dx)[(3/2) * x^2]
= (3/2) * (d/dx)(x^2)
= (3/2) * 2x
= 3x
We can see that g'(x) = 3x, which is the same as f(x). This confirms that g(x) is indeed the antiderivative of f(x).
The Fundamental Theorem of Calculus states that if F(x) is an antiderivative of a function f(x), then the definite integral of f(x) from a to b can be calculated as F(b) - F(a). In this case, since g(x) is the antiderivative of f(x), we can calculate the area under the graph of f(t) between 0 and x as g(x) - g(0) = g(x) - 0 = g(x).
Therefore, g(x) represents the area under the graph of f(t) between 0 and x, confirming Part 1 of the Fundamental Theorem of Calculus for the function f.
To find a formula for the function g(x), we need to determine the area under the graph of f(t) between 0 and x.
(a) To find the area under the graph of f(t) between 0 and x, we can calculate the area of the rectangle formed by the base of x and the height of f(x) at that point.
Considering the graph of f(t) = 3t, the height of the rectangle at any given point x is given by f(x) = 3x, since the y-coordinate represents the value of the function.
The base of the rectangle is x. Therefore, the area of the rectangle is given by A = base × height = x × f(x) = x × 3x = 3x^2.
Thus, the formula for g(x) is g(x) = 3x^2.
(b) To verify that g(x) is an antiderivative of f(x), we need to differentiate g(x) and check if it matches f(x).
Differentiating g(x) = 3x^2 with respect to x, we get dg(x)/dx = 6x.
The derivative of f(x) = 3x is df(x)/dx = 3.
Since dg(x)/dx = df(x)/dx = 3, the derivative of g(x) matches the function f(x). This confirms that g(x) is indeed an antiderivative of f(x).
This confirms Part 1 of the Fundamental Theorem of Calculus, which states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a).
In this case, g(x) is an antiderivative of f(x), and the area under the graph of f(x) between 0 and x is given by g(x), as shown in Part (a). Therefore, the Fundamental Theorem of Calculus holds true for the function f(x) = 3x.