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.Use the comparison principle and the intermediate value theorem

To prove the given statement, we will use the comparison principle and the intermediate value theorem.

Step 1: Define the function h(x) as given in the hint.

h(x) = 1/2 f(x)(b^2 - a^2) - ∫ from a to b f(t) dt

Step 2: Evaluate h(a) and h(b).

h(a) = 1/2 f(a)(b^2 - a^2) - ∫ from a to b f(t) dt h(b) = 1/2 f(b)(b^2 - a^2) - ∫ from a to b f(t) dt

Step 3: Since f is continuous on [a, b], there exists a minimum m and a maximum M such that m ≤ f(x) ≤ M for all x in [a, b].

Step 4: We can then say that 1/2 m(b^2 - a^2) ≤ ∫ from a to b f(t) dt ≤ 1/2 M(b^2 - a^2)

Step 5: This implies that h(a) ≤ 0 and h(b) ≥ 0.

Step 6: Now, we can apply the Intermediate Value Theorem, which states that if a function is continuous on a closed interval [a, b] and k is any number between f(a) and f(b), then there exists at least one number c in the interval (a, b) such that f(c) = k.

Step 7: Since h is continuous on [a, b] and h(a) ≤ 0 ≤ h(b), by the Intermediate Value Theorem, there exists a c in [a, b] such that h(c) = 0.

Step 8: Therefore, there exists a c in [a, b] such that 1/2 f(c)(b^2 - a^2) = ∫ from a to b f(t) dt, which is what we wanted to prove.

To prove the given statement, we will use the comparison principle and the intermediate value theorem.

Step 1: Define the function h(x) as given in the hint.

h(x) = 1/2 * f(x) * (b^2 - a^2) - ∫ from a to b f(t) dt

Step 2: Notice that h(a) = 1/2 * f(a) * (b^2 - a^2) - ∫ from a to b f(t) dt and h(b) = 1/2 * f(b) * (b^2 - a^2) - ∫ from a to b f(t) dt

Step 3: Since f is continuous on [a, b], the function h is also continuous on [a, b].

Step 4: By the comparison principle, we know that f(a) ≤ f(t) ≤ f(b) for all t in [a, b].

Step 5: Multiply the inequality by (b^2 - a^2)/2 and subtract ∫ from a to b f(t) dt from all parts of the inequality. We get h(a) ≤ h(t) ≤ h(b) for all t in [a, b].

Step 6: By the intermediate value theorem (which can be applied because h is continuous on [a, b]), there exists some c in [a, b] such that h(c) = 0.

Step 7: If we substitute h(c) = 0 into the definition of h(x), we get 1/2 * f(c) * (b^2 - a^2) - ∫ from a to b f(t) dt = 0.

Step 8: Rearranging the terms, we get ∫ from a to b f(t) dt = 1/2 * f(c) * (b^2 - a^2), which is the required result.

Therefore, there exists some c in [a, b] such that ∫ from a to b f(t) dt = 1/2 * f(c) * (b^2 - a^2).