Now compare the two results.How is the definite integral   related to F(10) – F(5)?

The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.(a)    10f(x) dx0 (b)    25f(x) dx0 (c)    35f(x) dx25 (d)    45f(x) dx0

To solve for \( \int_2^5 f(x) \, dx \) given that \( \int_0^5 f(x) \, dx = 3 \) and \( \int_0^2 f(x) \, dx = -2 \), follow these steps: 1. **Understand the relationship between the integrals**: The integral from 0 to 5 can be split into two parts: \[ \int_0^5 f(x) \, dx = \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \] 2. **Substitute the given values**: \[ 3 = \int_0^2 f(x) \, dx + \int_2^5 f(x) \, dx \] Given that \( \int_0^2 f(x) \, dx = -2 \), substitute this value into the equation: \[ 3 = -2 + \int_2^5 f(x) \, dx \] 3. **Solve for \( \int_2^5 f(x) \, dx \)**: \[ 3 = -2 + \int_2^5 f(x) \, dx \] \[ \int_2^5 f(x) \, dx = 3 + 2 \] \[ \int_2^5 f(x) \, dx = 5 \] So, the value of \( \int_2^5 f(x) \, dx \) is \( 5 \). The correct answer is: - \( 5 \)

Let f be a function such that f (1) = − 2 and f (5) = 7. Which of the following conditions ensures that f(c) = 0 for some value c in the open interval (1,5)?Responses∫15𝑓(𝑥)𝑑𝑥 exists. Integration of f of x of d x exists from 1 to 5f is increasing on the closed interval [1, 5].f is increasing on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is continuous on the closed interval [1, 5].f is defined for all values of x in the closed interval [1, 5].

EXAMPLE 4 Evaluate the following integrals by interpreting each in terms of areas.(a)    39 − x2 dx0(b)    10(x − 3) dx0SOLUTION (a) Since f(x) = 9 − x2 ≥ 0, we can interpret this integral as the area under the curve y = 9 − x2 from 0 to . But since y2 = , we get x2 + y2 = 9, which shows that the graph of f is a quarter-circle with radius in the top figure. Therefore,39 − x2 dx0 = 14𝜋(3)2 = .(b) The graph of y = x − 3 is the line with slope shown in the bottom figure. We compute the integral as the difference of the areas of the two triangles:10(x − 3) dx0 = A1 − A2 = − 4.5 = .

EXAMPLE 7 If it is known that 6f(x) dx0 = 10 and 2f(x) dx0 = 5, find 6f(x) dx2.SOLUTION By a property of integrals, we have2f(x) dx0 + 6f(x) dx2 = 6f(x) dx0so6f(x) dx2 =  6f(x) dx0 − 2f(x) dx0 =  10 −  =  .