10. Let 0 < a < b and f : [a, b] → R be continuous on [a, b]. Prove that there exists c ∈ [a, b] suchthatZ batf (t)dt = 12 f (c)(b2 − a2).Hint) Consider the functionh(x) = 12 f (x)(b2 − a2) −Z batf (t)dt

This problem is asking you to prove a specific property of a continuous function on a given interval. Here's a step-by-step solution:

1. Define the function h(x) as given in the problem: h(x) = 1/2 f(x)(b^2 - a^2) - ∫ from a to b of f(t) dt.

2. Since f is continuous on [a, b], by the Fundamental Theorem of Calculus, the integral of f from a to b is well-defined. Therefore, h(x) is also well-defined for all x in [a, b].

3. We want to show that there exists some c in [a, b] such that h(c) = 0.

4. Notice that h(a) = 1/2 f(a)(b^2 - a^2) - ∫ from a to b of f(t) dt and h(b) = 1/2 f(b)(b^2 - a^2) - ∫ from a to b of f(t) dt.

5. Since 0 < a < b, we know that (b^2 - a^2) > 0. Therefore, the sign of h(x) at x = a and x = b is determined by the sign of f(a) and f(b), respectively.

6. If either f(a) = 0 or f(b) = 0, then we have found our c such that h(c) = 0.

7. If f(a) and f(b) are both non-zero, then since f is continuous on [a, b], by the Intermediate Value Theorem, f must take on every value between f(a) and f(b) on the interval [a, b]. Therefore, there must exist some c in [a, b] such that f(c) = 0, and hence h(c) = 0.

8. Therefore, in all cases, we have shown that there exists some c in [a, b] such that h(c) = 0, as required.