Question

Let \(f(x)\) be continuous for \(a \leq x \leq b\). a) Prove that $$ \frac{d}{d x} \int_{x}^{b} f(t) d t=-f(x), \quad a \leq x \leq b . $$ b) Prove that $$ \frac{d}{d x} \int_{a}^{x^{2}} f(t) d t=2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b . $$ sit ? Thint: Let \(u=x^{2}\), so that the integral becomes a function \(F(u)\). Then, by the chain rule, \(\frac{d}{d x} F(u)=F^{\prime}(u) \frac{d u}{d x}\). ] c) Prove that $$ \frac{d}{d x} \int_{x^{2}}^{b} f(t) d t=-2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b . $$ d) Prove that $$ \frac{d}{d x} \int_{x^{2}}^{x^{3}} f(t) d t=3 x^{2} f\left(x^{3}\right)-2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b, \quad a \leq x^{3} \leq b . $$ [Hint: Let \(u=x^{2}, v=x^{3}\). Then the integral is \(F(u, v)\). By the chain rule, $$ \left.\frac{d}{d x} F(u, v)=\frac{\partial F}{\partial u} \frac{d u}{d x}+\frac{\partial F}{\partial v} \frac{d v}{d x} .\right] $$

Short answer

In summary, we have successfully proven the following statements about the derivatives of integrals concerning different lower and upper bounds: a) The derivative of an integral with the upper bound x is equal to -f(x). b) The derivative of an integral with the upper bound x^2 is equal to 2xf(x^2). c) The derivative of an integral with the lower bound x^2 is equal to -2xf(x^2). d) The derivative of an integral with lower bound x^2 and upper bound x^3 is equal to 3x^2f(x^3)-2xf(x^2). These proven statements help to further understand the properties of derivatives and how they apply to integrals in different situations.

Step-by-step solution

Part a: Prove Derivative of Integral with Upper Bound x

To prove the statement $$ \frac{d}{d x} \int_{x}^{b} f(t) d t=-f(x), \quad a \leq x \leq b, $$ we need to find the derivative of the integral with respect to its lower bound, x. Using the Fundamental Theorem of Calculus, it states that if \(g(x) = \int_{x}^{b} f(t) dt\), then $$ g'(x) = -f(x). $$ By differentiating the integral, we obtain the desired result.

Part b: Prove Derivative of Integral with Upper Bound x^2

To prove the statement $$ \frac{d}{d x} \int_{a}^{x^{2}} f(t) d t=2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b, $$ we'll perform a substitution and use the Chain Rule. Let \(u=x^2\), so the integral becomes \(F(u) = \int_{a}^{u} f(t) dt\). Now, by the Chain Rule, $$ \frac{d}{d x} F(u) = F'(u) \frac{d u}{d x} = 2xf(x^2). $$ Therefore, the derivative of the integral with respect to x is found to be the desired result.

Part c: Prove Derivative of Integral with Lower Bound x^2

To prove the statement $$ \frac{d}{d x} \int_{x^{2}}^{b} f(t) d t = -2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b, $$ we once again perform a substitution and apply the Chain Rule. Let \(u=x^2\), so the integral becomes \(F(u) = \int_{u}^{b} f(t) dt\). By the Chain Rule, $$ \frac{d}{d x} F(u) = F'(u) \frac{d u}{d x}=-2xf(x^2). $$ This result matches the desired result in the statement.

Part d: Prove Derivative of Integral with Bounds x^2 and x^3

To prove the statement $$ \frac{d}{d x} \int_{x^{2}}^{x^{3}} f(t) d t=3 x^{2} f\left(x^{3}\right)-2 x f\left(x^{2}\right), \quad a \leq x^{2} \leq b, \quad a \leq x^{3} \leq b, $$ we'll perform substitutions and use the Chain Rule for multivariable functions. Let \(u=x^2\) and \(v=x^3\). Then the integral is \(F(u, v) = \int_{u}^{v} f(t) dt\). By the Chain Rule, $$ \frac{d}{d x} F(u, v)=\frac{\partial F}{\partial u} \frac{d u}{d x}+\frac{\partial F}{\partial v} \frac{d v}{d x}. $$ Applying the Chain Rule, we get $$ \frac{d}{d x} F(u, v) = -f(u) \cdot 2x + f(v) \cdot 3x^2 = 3 x^2 f\left(x^{3}\right) - 2 x f\left(x^{2}\right). $$ So the derivative of the integral with respect to x has been found to be the desired result.

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus, essential when dealing with composite functions, where one function is applied inside another. If you imagine a function like a machine that transforms an input into an output, a composite function is a series of machines working together: the output from one machine becomes the input for the next.
In more technical terms, the chain rule helps find the derivative of such compositions. Let's say you have a function \(y\) which is a composite of two functions \(u\) and \(v\). If \(y = f(u)\) and \(u = g(v)\), then the derivative \(\frac{dy}{dv}\) is found using:

• \(\frac{dy}{dv} = \frac{dy}{du} \times \frac{du}{dv}\)

This rule is woven through problems of derivatives of integrals when the limits themselves are functions of \(x\).
Here's how it appears: when differentiating an integral like \(\int_{a}^{x^2} f(t) \, dt\), first we identify the function \(u = x^2\), transforming the problem into differentiation of \(F(u) = \int_{a}^{u} f(t) \, dt\). The chain rule then steps in to adjust for this inner change, multiplying the result by the derivative of \(u\), which is \(2x\). This precision is crucial, ensuring the correct transformation as inputs change.

Derivative of Integrals

The Fundamental Theorem of Calculus is key when understanding the derivative of integrals. This theorem bridges the concept of antiderivatives (integrals) and derivatives, showing how they are, in essence, reverses of each other.
Consider an integral with bounds \(\int_{a}^{x} f(t) \, dt\). If we denote the integral as a function \(F(x)\), then the variety of \(F\) can be virtually undone by taking its derivative: \(\frac{d}{dx}F(x) = f(x)\). This demonstrates that integration and differentiation are inverse operations.
In more complex scenarios, however, the bounds themselves may be functions of \(x\). This introduces the necessity to take the derivative of the integral starting from an inner or outer bound, as various parts can impact the result.
For example:

• When starting in \(x^2\) rather than \(a\) or \(x\), the derivative \(\frac{d}{dx}\int_{x^2}^{b} f(t) \, dt\) reflects this shift, yielding \(-2x f(x^2)\). This entails both understanding \(f(t)\)'s behavior and adjusting for the \(x^2\) factor from the chain rule. It's a potent demonstration of calculus's reach, neatly folding together past topics into one complete concept.

Remember, the differentiation operation reverses the integration up to the function represented by the integrand at the upper limit of the integration.

Continuous Functions

Continuity in functions plays a central role in calculus. A function \(f(x)\) is continuous on an interval if there are no breaks, jumps, or holes over that segment, meaning you can draw it without lifting your pen from the paper.
Mathematically, this means that as \(x\) approaches a point \(c\) in the interval, \(f(x)\) approaches \(f(c)\). In simpler terms:

• \(\lim_{x \to c} f(x) = f(c)\)

Continuity ensures the Fundamental Theorem of Calculus holds because we rely on the unbroken nature of \(f\) over \([a, b]\).
In integrals:

• The bounds \(x^2\) or \(x^3\) introduce continuity conditions. When we shift these limits not just to \(a\) and \(x\) but to \(x^2\) and \(x^3\), continuity gives the assurance that resulting expressions, like \(f(x^2)\) and \(f(x^3)\), are well-defined.

Without continuity, you would face undefined or infinite results, which mathematically and logically, aren't useful for precise calculation. Thus, each problem involving derivatives of integrals starts by ensuring the function at hand is continuous.

Multivariable Calculus

In multivariable calculus, problems involve functions with multiple inputs, or variables, making things less straightforward but very interesting. Much like how single-variable calculus explores the rate of change in one dimension, multivariable calculus broadens this to explore many dimensions at once.
When examining integrals with bounds that aren't constants, but instead functions upon themselves like \(x^2\) or \(x^3\), it's akin to working in this multi-dimension because each bound itself depends on \(x\).
In these scenarios, the derivative involves not a single change but multiple interactions:

• For instance, while handling the integral \(\int_{x^2}^{x^3} f(t) dt\), you see the fundamental theorem extend to\(\frac{d}{dx}\), managing changes at both limits.
• Each part, \(u = x^2\) and \(v = x^3\), contributes through \(\frac{\partial F}{\partial u}\) and \(\frac{\partial F}{\partial v}\), with derivatives \(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = 3x^2\).
• This results in a more nuanced calculation: \(-f(x^2) \cdot 2x + f(x^3) \cdot 3x^2\).

Understanding these layers of operations broadens your understanding and mastery of calculus, preparing you for more complex mathematical challenges.