Question

Prove that \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{\mathrm{a}}^{g(x)} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\mathrm{f}[\mathrm{g}(\mathrm{x})] \mathrm{g}^{\prime}(\mathrm{x})\).

Short answer

Question: Prove that \(\frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\). Answer: We followed a step-by-step solution using the definition of the derivative and the Fundamental Theorem of Calculus Part 1 (FTC-1). We proved that \(\frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\) by applying the properties of integrals, FTC-1, and the chain rule.

Step-by-step solution

Start with the definition of the derivative

We will find the derivative of the given integral with respect to x using the definition of the derivative: \(\frac{\mathrm{d}}{\mathrm{dx}} F(x) =\lim_{h\to 0} \frac{F(x+h)-F(x)}{h}\).

Replace F(x) with the given integral

Let's let \(F(x)=\int_{a}^{g(x)} f(t) dt\). Then, the expression for the derivative becomes: \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{a}^{g(x)} f(t) dt = \lim_{h\to 0} \frac{\int_{a}^{g(x+h)} f(t) dt - \int_{a}^{g(x)} f(t) dt}{h}\).

Apply the properties of integrals

The difference of two integrals with the same integrand and the same lower bound can be written as a single integral. In this case, we can rewrite the limit as: \(\lim_{h\to 0} \frac{\int_{a}^{g(x+h)} f(t) dt - \int_{a}^{g(x)} f(t) dt}{h} = \lim_{h\to 0} \frac{\int_{g(x)}^{g(x+h)} f(t) dt}{h}\).

Apply the Fundamental Theorem of Calculus Part 1 (FTC-1)

According to FTC-1, if \(f\) is continuous on the interval \([g(x), g(x+h)]\), then there exists a function \(F\) such that \(F'(x)=f(x)\). So, we can rewrite the expression inside the limit as: \(\int_{g(x)}^{g(x+h)} f(t) dt = F(g(x+h))-F(g(x))\). Next, substitute the expression back into the limit: \(\lim_{h\to 0} \frac{F(g(x+h))-F(g(x))}{h}\).

Apply the chain rule

Recall from calculus that the chain rule states that if \(F\) and \(g\) are differentiable functions, then the derivative of \(F(g(x))\) with respect to x is given by: \(\frac{d}{dx}(F(g(x)))=F'(g(x))g'(x)\). Using this fact, we can rewrite the limit as: \(\lim_{h\to 0} \frac{F(g(x+h))-F(g(x))}{h} = F'(g(x))g'(x)\).

Prove the given expression

Recall from Step 4 that we have: \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{a}^{g(x)} f(t) dt = \lim_{h\to 0} \frac{\int_{g(x)}^{g(x+h)} f(t) dt}{h}\). After applying the chain rule, we have: \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{a}^{g(x)} f(t) dt = F'(g(x))g'(x)\). Finally, substitute \(F'(x)=f(x)\) based on FTC-1, and we get: \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{a}^{g(x)} f(t) dt = f(g(x)) g'(x)\). We have now proven that \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{a}^{g(x)} f(t) dt = f(g(x)) g'(x)\).

Key concepts

Definite Integrals

A definite integral is a fundamental concept in calculus representing the area under a curve, defined for function f(t) from a point a to a point b. It's represented by the notation \[\begin{equation}\text{\(\int_{a}^{b} f(t) \text{dt}\)}\end{equation}\]. The value of a definite integral is a number that describes the accumulation of quantities, such as area, volume, or mass, given the density distribution described by the function \(f\).

Definite integrals adhere to several properties. For example, if the upper and lower bounds are the same, the integral equals zero. Another important property is that they can be additive; the integral from \(a\) to \(c\), plus the integral from \(c\) to \(b\), is equal to the integral from \(a\) to \(b\). Understanding these properties helps when solving more complex problems—like when the upper limit is a function of \(x\) as seen with \(g(x)\) in our main exercise.

To solve the problem in our exercise, we recognize the integrals’ capacity to represent continuous sums and how we can operate on them to find a derivative, which leads us to the next core concept: the derivative of an integral.

Derivative of an Integral

The derivative of an integral may initially seem perplexing, but it's a powerful tool in calculus known as the Fundamental Theorem of Calculus (FTC). This theorem bridges the two primary operations in calculus: differentiation and integration.

The FTC states that if \( f(t) \) is continuous over an interval containing \(a\) and \(g(x)\), then the function \(F(x)\), defined by
\[\begin{equation}\(F(x) = \int_{a}^{g(x)} f(t) dt\)\end{equation}\]
, has a derivative and that derivative is given by \( f \) evaluated at the upper limit of integration:
\[\begin{equation}\frac{d}{dx}F(x) = f(g(x)) \bullet g'(x)\end{equation}\]

The process of taking the derivative of an integral involves this specific instance of the Fundamental Theorem of Calculus, which has a clear application in our exercise's solution. By recognizing the integral as a function with variable upper limits, we can differentiate using traditional methods, as demonstrated through the problem's step-by-step breakdown.

Chain Rule

The chain rule is another cornerstone of calculus, particularly relevant when dealing with composite functions—or functions within functions. Mathematically, if we have two functions \(F\) and \(g\), where \(F\) is a function of \(u=g(x)\), the chain rule tells us how to find the derivative of the composite function \(F(g(x))\).

The rule states that:
\[\begin{equation}\frac{d}{dx}[F(g(x))] = F'(g(x)) g'(x)\end{equation}\]

This is invaluable when evaluating the derivative of an integral with a variable upper limit. As the numerator in our main exercise depends on \(x\), applying the chain rule allows us to differentiate \(F(g(x+h)) - F(g(x))\) without directly tackling the complexity of \(F\) itself. Our step-by-step solution leverages this rule to simplify the expression, ultimately proving the initial problem correct. The ability to break down complex functions into their components is key to successfully applying calculus concepts, especially in exercises requiring the derivative of an integral.