Understanding the Derivative of the Integral Function using the Fundamental Theorem of Calculus: Explained with Examples and Step-by-Step Derivation

If h(x)=∫x3−12+t2−−−−−√ⅆt for x≥0, then h′(x)=

To find the derivative of the function h(x), we can use the Fundamental Theorem of Calculus

To find the derivative of the function h(x), we can use the Fundamental Theorem of Calculus. According to this theorem, if a function F(x) is an antiderivative of f(x), then the derivative of the integral of f(x) from a constant a to x is given by f(x).

In our case, the function h(x) is defined as the integral of x^3 – 12 + t^2√(dt) from a constant 0 to x. Let’s rewrite the integral as:

h(x) = ∫[0 to x] (x^3 – 12 + t^2√(dt))

Now, let’s use the Fundamental Theorem of Calculus to find the derivative of h(x).

Since the upper bound of the integral is x, we can treat it as a variable and differentiate with respect to x. The lower bound, 0, is a constant and does not contribute to the derivative.

Thus, applying the Fundamental Theorem of Calculus, we have:

h'(x) = d/dx ∫[0 to x] (x^3 – 12 + t^2√(dt))

To find the derivative, we need to apply the chain rule. The chain rule states that if we have a composite function g(f(x)), then the derivative of g(f(x)) is given by g'(f(x)) * f'(x).

In our case, we have the composite function g(f(x)) = ∫[0 to f(x)] (x^3 – 12 + t^2√(dt)). The outer function g(u) is the integral and the inner function f(x) = x.

Let’s differentiate g(f(x)) using the chain rule:

h'(x) = g'(f(x)) * f'(x)

The derivative of the outer function g(u) = ∫[0 to u] (x^3 – 12 + t^2√(dt)) with respect to u is simply the integrand evaluated at u. Since f(x) = x, we substitute u = x in the integrand:

g'(f(x)) = (x^3 – 12 + x^2√) evaluated at u = f(x) = x

g'(f(x)) = (x^3 – 12 + x^2√) evaluated at x = x

Now, let’s differentiate the inner function f(x) = x with respect to x:

f'(x) = 1

Substituting these values back into the chain rule equation, we get:

h'(x) = (x^3 – 12 + x^2√) evaluated at x = x * 1

Therefore, the derivative of h(x) is:

h'(x) = x^3 – 12 + x^2√

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