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? ​

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 −  =  .

To find \(\int_{-1}^{1} f(x) \, dx\), we can use the given integrals and the properties of definite integrals. We know: \[ \int_{-4}^{5} f(x) \, dx = -8.0 \] \[ \int_{-4}^{-1} f(x) \, dx = -2.4 \] \[ \int_{1}^{5} f(x) \, dx = -6.1 \] We can break down the integral \(\int_{-4}^{5} f(x) \, dx\) into parts: \[ \int_{-4}^{5} f(x) \, dx = \int_{-4}^{-1} f(x) \, dx + \int_{-1}^{1} f(x) \, dx + \int_{1}^{5} f(x) \, dx \] Substitute the known values: \[ -8.0 = -2.4 + \int_{-1}^{1} f(x) \, dx + (-6.1) \] Combine the constants: \[ -8.0 = -2.4 - 6.1 + \int_{-1}^{1} f(x) \, dx \] \[ -8.0 = -8.5 + \int_{-1}^{1} f(x) \, dx \] Solve for \(\int_{-1}^{1} f(x) \, dx\): \[ \int_{-1}^{1} f(x) \, dx = -8.0 + 8.5 \] \[ \int_{-1}^{1} f(x) \, dx = 0.5 \] Therefore, the correct answer is: \[ \boxed{C} \]

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) =

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].

If f(4) = 11, f ' is continuous, and 6f '(x) dx4 = 18, what is the value of f(6)?f(6) =