Problem 2

Question

Compute $\frac{d}{d x}\left(\int_{0}^{x^{2}} \sin \left(s^{2}\right) d s\right)$.

Step-by-Step Solution

Verified

Answer

The derivative is $2x \sin(x^{4})$.

1Step 1: Understand the Problem

We are asked to find the derivative of an integral with respect to the upper limit, which is a function of $x$. This is a problem suitable for the Leibniz rule or the Fundamental Theorem of Calculus.

2Step 2: Utilize the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) part 2 states that if $ F(b) = \int_{a}^{b} f(s) \, ds $, then the derivative of this integral with respect to $ b $ is $ f(b) $. For a simple upper limit, this would be $ \sin(x^2) $. However, our problem involves a variable upper limit $x^2$.

3Step 3: Apply the Chain Rule

Since the upper limit is $x^2$ instead of $ x $, we need to apply the chain rule. Let $ u = x^2 $, so $ \frac{d}{dx}\int_{0}^{u}\sin(s^2) \, ds = \sin(u^2) \cdot \frac{d}{dx}(u) $.

4Step 4: Calculate the Derivative

Now calculate $ \frac{d}{dx} (x^2) = 2x $. Hence, the derivative becomes $ \sin((x^2)^2) \cdot 2x = 2x \sin(x^4) $.

5Step 5: Final Result

Thus, the derivative $ \frac{d}{d x}\left(\int_{0}^{x^{2}} \sin(s^{2}) ds \right) = 2x \sin(x^4) $.

Key Concepts

Leibniz rulechain rulederivatives of integrals

Leibniz rule

The Leibniz rule is a powerful method used in calculus when dealing with derivatives of integrals, especially when the limits of integration are not constants but functions of the variable we are differentiating. This rule allows us to find the derivative of such integrals efficiently.
In mathematical terms, if you have an integral of the form $ \int_{a(x)}^{b(x)} f(s) \, ds $, the derivative with respect to $ x $ is given by:

$ f(b(x)) \cdot \frac{d}{dx} b(x) $ minus $ f(a(x)) \cdot \frac{d}{dx} a(x) $
Plus the integral of the partial derivative of the integrand with respect to $ x $.

In our specific problem, the integral $ \int_{0}^{x^2} \sin(s^2) \, ds $ only has a variable in the upper limit $ b(x) = x^2 $, while the lower limit $ a(x) = 0 $ is constant. Thus, we simplify the use of the Leibniz rule to just consider the change in the upper limit.
This results in using the Fundamental Theorem of Calculus combined with the Chain Rule, which then provides a path to solving the given problem.

chain rule

The chain rule is an essential principle in calculus that aids in differentiating composite functions. When a variable depends on another variable indirectly through a function, we use the chain rule.
Here's a simple way to understand it: if you have a function $ y = f(u) $ and $ u = g(x) $, the derivative $ \frac{dy}{dx} $ can be found using:

Take the derivative of $ f $ with respect to $ u $: $ \frac{dy}{du} $.
Then multiply it by the derivative of $ u $ with respect to $ x $: $ \frac{du}{dx} $.

In our original exercise, after identifying that the upper limit of our integral is an expression $ u = x^2 $, the chain rule becomes crucial. We first differentiate the inner function $ u $, yielding $ \frac{du}{dx} = 2x $.
So, while differentiating an integral with respect to its upper limit that is itself a function of $ x $, recognizes this nested relationship, the chain rule dictates multiplying the derivative of the outer function by the derivative of the inner function.

derivatives of integrals

Finding the derivative of an integral shares an intricate yet fascinating connection through the Fundamental Theorem of Calculus. When given an integral with a variable upper limit, say $ \int_{0}^{u} f(s) \, ds $, the theorem states that the derivative of this integral with respect to $ u $ is simply $ f(u) $.
However, if the upper limit itself is a function of another variable, as in our exercise where $ u = x^2 $, we approach the problem by considering both this theorem and the chain rule.

The integral $ \int_{0}^{x^2} \sin(s^2) \, ds $ demonstrates this relationship elegantly.
By applying the Fundamental Theorem, we recognize the derivative $ f(u) = \sin((x^2)^2) $.
The functionality of the chain rule allows us to include the derivative $ \frac{d}{dx} (x^2) = 2x $.

Ultimately, these two concepts together result in $ 2x \sin(x^4) $, illustrating how intricately derivatives and integrals are woven together in calculus, allowing us to handle even complex compositions like those seen in the problem at hand.

Problem 2

Problem 3

Other exercises in this chapter

Problem 2

Find out for which $a \in \mathbb{R}$ does $\sum_{n=1}^{\infty} e^{a n}$ converge. When the series converges, find an upper bound for the sum.

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Problem 2

Let $f:[0,1] \rightarrow \mathbb{R}$ be defined by $f(x):=x .$ Show that $f \in \mathscr{A}[0,1]$ and compute $\int_{0}^{1} f$ using the definition of t

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Problem 3

(requires \$4.3): Use Taylor's theorem to study the remainder term and show that for all $x \in \mathbb{R}$ $$ e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} $$

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Problem 3

Let $f:[a, b] \rightarrow \mathbb{R}$ be Riemann integrable. Let $g:[a, b] \rightarrow \mathbb{R}$ be a function such that $f(x)=$ $g(x)$ for all \(x \i

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