Question

Suppose h is a function such that h(2) = 2, h'(2) = 2, h''(2) = 4, h(4) = 7, h'(4) = 6, h''(4) = 12, and h' is continuous everywhere. Evaluate h''(u) du.

Find the derivative of the function below:
g(c) = ∫ (5 + t²) dt
Since f(t) = √(5 + t⁶) is continuous, Part 1 of the Fundamental Theorem of Calculus gives the answer:
g'(2)

Answer 1

Integral of h''(u) without limits is h'(u) plus a constant. The derivative of g(c) = ∫ (5 + t²) dt is g'(c) = 2c. For f(t) = √(5 + t⁶), g'(2) is evaluated using the Fundamental Theorem of Calculus Part 1, resulting in 96.

Step-by-step explanation:

To evaluate the integral of h''(u) with respect to u, which is represented as §h''(u) du, we must have the limits of integration or additional context to find a specific value. However, since the integral is simply the antiderivative of h''(u), without limits, the result will be h'(u) plus a constant of integration, assuming h'(u) is continuous everywhere as stated.

The function g(c) represents the integral of the expression 5 + t² with respect to t. To find the derivative g'(c), by the Fundamental Theorem of Calculus Part 1, we would differentiate the integrand 5 + t² with respect to t and then evaluate it at t = c. Thus, g'(c) = 2c.

For f(t) = √(5 + t⁶), using the Fundamental Theorem of Calculus Part 1, the derivative of g(c) at c = 2 is evaluated by substituting the value into the derivative of the function inside the square root, multiplied by the derivative of the square root function itself, yielding g'(2) = 3 * 2^5 = 96.