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 −  =  .

Let ff be a continuous function such that integral, from, 1, to, 6, of, f, of, x, d, x, equals, 2∫ 16​ f(x)dx=2 and integral, from, 6, to, minus, 6, of, f, of, x, d, x, equals, 8, .∫ 6−6​ f(x)dx=8. What is the value of integral, from, minus, 6, to, 1, of, f, of, x, d, x, question mark∫ −61​ f(x)dx? ​

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: - \( 5 \)

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)f(x) = 6x5 − 7x4 − 9x2F(x) =

If 10f(x) dx = 390 and 10g(x) dx = 16,0 find 10[4f(x) + 6g(x)] dx.0

Find 7f(x) dx0 iff(x) = 5     for  x < 5x     for  x ≥ 5.