Derivative Of Integral: A Step-by-Step Calculus Guide

by Omar Yusuf 54 views

Hey guys! Today, we're diving deep into the fascinating world of calculus, tackling a problem that combines the power of integration and differentiation. Our mission? To compute the derivative of an integral. Specifically, we want to find:

ddx(∫−xx es2ds)\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right)

This might look a bit intimidating at first, but don't worry, we'll break it down step by step and make sure you understand the concepts involved. So, buckle up and let's embark on this calculus expedition!

Decoding the Problem: Fundamental Theorem of Calculus to the Rescue

Before we jump into the solution, let's take a moment to understand the core concept that will guide us: the Fundamental Theorem of Calculus (FTC). This theorem is the cornerstone of calculus, linking differentiation and integration in a profound way. There are actually two parts to the FTC, but the one we'll be using most directly here is the first part.

The first part of the FTC essentially states that if we have a function defined as an integral with a variable upper limit, then the derivative of that function is simply the integrand evaluated at the upper limit. In simpler terms, if we have:

F(x)=∍axf(t)dtF(x) = \int_a^x f(t) dt

where a is a constant, then:

F′(x)=f(x)F'(x) = f(x)

This is a super powerful result! It tells us that differentiation and integration are, in a sense, inverse operations. But wait, our problem has a twist! We have variable limits of integration, both x and -x. So, how do we handle that? That's where a little clever manipulation and another important rule, the chain rule, come into play.

Breaking Down the Integral: A Strategic Split

The key to tackling this problem is to split the integral into two parts. Why? Because the FTC in its simplest form applies when we have a constant lower limit of integration. We can introduce a constant, say 0, as a splitting point. This gives us:

∫−xx es2ds=∫−x0es2ds+∫0xes2ds\int ^x_{-x}\:e^{s^2}ds = \int_{-x}^0 e^{s^2} ds + \int_0^x e^{s^2} ds

Now we have two integrals, each with one variable limit and one constant limit. This is progress! But we're not quite there yet. Notice that the first integral has the variable limit in the lower bound, not the upper bound. To fix this, we can use a property of definite integrals: swapping the limits of integration changes the sign of the integral.

∫−x0es2ds=−∫0−xes2ds\int_{-x}^0 e^{s^2} ds = -\int_0^{-x} e^{s^2} ds

Substituting this back into our expression, we get:

∫−xx es2ds=−∫0−xes2ds+∫0xes2ds\int ^x_{-x}\:e^{s^2}ds = -\int_0^{-x} e^{s^2} ds + \int_0^x e^{s^2} ds

Alright, we've successfully massaged our integral into a form that's much more manageable. Now we can apply the FTC and the chain rule to find the derivative.

Unleashing the Derivative: FTC and the Chain Rule in Action

Now comes the exciting part: taking the derivative! Let's differentiate both sides of our equation with respect to x:

ddx(∫−xx es2ds)=ddx(−∫0−xes2ds+∫0xes2ds)\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = \frac{d}{dx}\left(- \int_0^{-x} e^{s^2} ds + \int_0^x e^{s^2} ds\right)

We can differentiate the sum term by term:

ddx(∫−xx es2ds)=−ddx(∫0−xes2ds)+ddx(∫0xes2ds)\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = -\frac{d}{dx}\left( \int_0^{-x} e^{s^2} ds\right) + \frac{d}{dx}\left(\int_0^x e^{s^2} ds\right)

For the second term, the FTC Part 1 applies directly! The derivative of ∍0xes2ds\int_0^x e^{s^2} ds with respect to x is simply ex2e^{x^2}.

But for the first term, we need to be a bit more careful. We have a function of a function inside the integral limit (-x), so we'll need to use the chain rule. Let's define a new function:

u(x)=−xu(x) = -x

Then the first term becomes:

−ddx(∫0u(x)es2ds)-\frac{d}{dx}\left( \int_0^{u(x)} e^{s^2} ds\right)

By the chain rule, this is equal to:

−ddu(∫0ues2ds)⋅dudx-\frac{d}{du}\left( \int_0^{u} e^{s^2} ds\right) \cdot \frac{du}{dx}

Now we can apply the FTC Part 1 to the first factor, which gives us eu2e^{u^2}. The second factor, dudx\frac{du}{dx}, is simply the derivative of -x with respect to x, which is -1. So, putting it all together, we get:

−eu2⋅(−1)=eu2-e^{u^2} \cdot (-1) = e^{u^2}

Substituting back u(x) = -x, we have:

e(−x)2=ex2e^{(-x)^2} = e^{x^2}

Now we have the derivatives of both parts of our original expression! Let's plug them back in:

ddx(∫−xx es2ds)=−ex2+ex2\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = -e^{x^2} + e^{x^2}

Wait a minute... it seems there was a sign error in the previous calculation. Let's correct it. The derivative of the first term should be:

−ddx(∫0−xes2ds)=−[e(−x)2⋅(−1)]=ex2- \frac{d}{dx} \left( \int_0^{-x} e^{s^2} ds \right) = - [e^{(-x)^2} \cdot (-1)] = e^{x^2}

So, the correct expression is:

ddx(∫−xx es2ds)=ex2+ex2\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = e^{x^2} + e^{x^2}

The Grand Finale: Simplifying to the Solution

We're almost there! We have:

ddx(∫−xx es2ds)=ex2+ex2\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = e^{x^2} + e^{x^2}

This simplifies beautifully to:

ddx(∫−xx es2ds)=2ex2\frac{d}{dx}\left(\int ^x_{-x}\:e^{s^2}ds\right) = 2e^{x^2}

And there you have it! We've successfully computed the derivative of the integral. The answer is 2ex22e^{x^2}.

Key Takeaways: Mastering the Calculus Concepts

Let's recap the key concepts we used to solve this problem:

  • Fundamental Theorem of Calculus (Part 1): This theorem is crucial for differentiating integrals with variable limits. It tells us that the derivative of an integral with a variable upper limit is the integrand evaluated at that limit.
  • Chain Rule: When we have a function of a function (like -x inside the integral limit), the chain rule is our best friend. It allows us to differentiate composite functions step by step.
  • Properties of Definite Integrals: Swapping the limits of integration changes the sign of the integral. This property is often helpful for manipulating integrals into a more convenient form.
  • Splitting the Integral: When dealing with variable limits of integration, splitting the integral at a constant point can make the problem easier to handle. This allows us to apply the FTC more directly.

By mastering these concepts, you'll be well-equipped to tackle a wide range of calculus problems involving differentiation and integration.

Practice Makes Perfect: Sharpening Your Calculus Skills

Now that we've conquered this problem, it's time to put your skills to the test! Here are a few similar problems you can try:

  1. Compute ddx(∫x2x sin(t2)dt)\frac{d}{dx}\left(\int ^x_{x^2}\:sin(t^2)dt\right).
  2. Find ddx(∫cos(x)sin(x) e−t2dt)\frac{d}{dx}\left(\int ^{sin(x)}_{cos(x)}\:e^{-t^2}dt\right).
  3. Calculate ddx(x∫0x e−t2dt)\frac{d}{dx}\left(x \int ^x_0\:e^{-t^2}dt\right).

Working through these problems will help solidify your understanding of the concepts we've discussed and build your confidence in tackling more complex calculus challenges.

Remember, calculus is a journey of exploration and discovery. Don't be afraid to make mistakes, ask questions, and keep practicing. With dedication and perseverance, you'll unlock the power of calculus and its many applications in the world around us. Keep exploring, guys! Happy calculating!