Question

question 7 2 pts a graph of f(x) is shown below, where f(x) is some antiderivative of a function f(x). suppose \\(\int_{0}^{k} f(t) \\, dt = 2\\). find the value of \\(k\\)

Explanation:

Step1: Recall the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \). In this case, \( a = 0 \) and \( b = k \), so \( \int_{0}^{k} f(t) \, dt = F(k) - F(0) \).

Step2: Determine \( F(0) \) from the graph

From the graph, when \( x = 0 \), \( F(0) = 1 \) (since the graph starts at \( (0, 1) \)).

Step3: Set up the equation and solve for \( k \)

We know that \( \int_{0}^{k} f(t) \, dt = 2 \). Using the Fundamental Theorem of Calculus, this becomes \( F(k) - F(0) = 2 \). Substituting \( F(0) = 1 \), we get \( F(k) - 1 = 2 \), so \( F(k) = 3 \). Now, we look at the graph of \( F(x) \) to find the \( x \)-value (which is \( k \)) where \( F(x) = 3 \). From the graph, when \( F(x) = 3 \), \( x = 9 \) (by looking at the grid and the curve, we can see that at \( x = 9 \), the function value is 3).

Answer:

\( 9 \)