# Derivative of integral bounded by functions

• springo
In summary, the derivative of the first integral is v'(x) * f ∘ v(x) - u'(x) * f ∘ u(x), and the derivative of the second integral is h'(x) * f ∘ h(x). These are applications of the Fundamental Theorem of Calculus, as shown through the use of antiderivatives and the chain rule.
springo

## Homework Statement

What's the derivative of the following two:
$$\int_{a}^{h(x)}f(t)\,\mathrm{d}t$$
$$\int_{u(x)}^{v(x)}f(t)\,\mathrm{d}t$$

## The Attempt at a Solution

I thought of doing the following:
$$\int_{h(a)}^{h(x)}f(t)\,\mathrm{d}t = \int_{a}^{x}f\circ h(u)\cdot h'(u)\,\mathrm{d}u$$
(with t = h(u) )
But then I don't know how to continue.

If the antiderivative of f(t) is F(t), then the integral of the second one is F(v(x))-F(u(x)), right? Differentiate that using the chain rule.

These problems are applications of the Fundamental Theorem of Calculus. As a hint, I'll work a different, but related, problem.

$$\int_a^{x^2} f(t) dt$$
Suppose an antiderivative of f(t) is F(t). I.e., F'(t) = f(t).
So
$$\int_a^{x^2} f(t) dt = F(x^2) - F(a)$$
If we differentiate both sides of this equation with respect to x, we get
$$d/dx \int_a^{x^2} f(t) dt = d/dx(F(x^2) - F(a))$$
= F'(x^2) * d/dx(x^2) - 0 (F(a) is a constant, so its derivative is zero)
= f'(x^2) * 2x

Is that enough to get you going?

OK, so I guess then...
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{u(x)}^{v(x)}f(t)\,\mathrm{d}t \right)=v'(x) \cdot f \circ v(x) - u'(x) \cdot f \circ u(x)$$
As for the other one, since u'(x) = da/dx = 0, we have:
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{a}^{h(x)}f(t)\,\mathrm{d}t \right)= h'(x)\cdot f\circ h(x)$$

Is that all OK?
Thanks.

Looks ok to me.

## 1. What is the meaning of the derivative of an integral bounded by functions?

The derivative of an integral bounded by functions is a mathematical concept that represents the rate of change of the area under the curve defined by the functions. It is also known as the rate of change of the net accumulation of the function values.

## 2. How is the derivative of an integral bounded by functions calculated?

The derivative of an integral bounded by functions can be calculated using the Fundamental Theorem of Calculus, which states that the derivative of an integral is equal to the integrand evaluated at the upper bound of the integral.

## 3. What is the relationship between the derivative and the integral bounded by functions?

The derivative and the integral bounded by functions are inverse operations of each other. This means that the derivative of an integral is equal to the original function, while the integral of a derivative is equal to the original function plus a constant.

## 4. Can the derivative of an integral bounded by functions be negative?

Yes, the derivative of an integral bounded by functions can be negative. This occurs when the upper bound of the integral is decreasing, resulting in a negative rate of change for the area under the curve.

## 5. What are some real-life applications of the derivative of an integral bounded by functions?

The derivative of an integral bounded by functions has many applications in fields such as physics, engineering, and economics. For example, it can be used to calculate the rate of change of displacement, velocity, and acceleration in motion, as well as the rate of change of quantities in a changing system. It is also used to solve optimization problems and determine maximum and minimum values of functions.

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