Definite Integral: Solving ∫6xe^(x^2) Dx (0 To 2)

Dec 2, 2025 by Andrew McMorgan 50 views

Evaluating the Definite Integral of 6xe^(x^2) from 0 to 2

Hey guys! Let's dive into the world of calculus and tackle a fun problem together. Today, we're going to break down how to evaluate the definite integral of 6xe^(x2) dx from the limits of 0 to 2. This type of problem might seem a bit intimidating at first, but with a step-by-step approach and a little trick called u-substitution, we'll make it super easy to understand. So, grab your calculators (or just your thinking caps!), and let's get started!

Understanding the Integral

Before we jump into solving, let's quickly recap what a definite integral actually represents. In simple terms, a definite integral calculates the area under a curve between two specified points on the x-axis. These points are what we call the limits of integration. In our case, we want to find the area under the curve of the function f(x) = 6xe^(x2) between x = 0 and x = 2. This area gives us a precise numerical value, unlike indefinite integrals which give us a general family of functions. Integrals are fundamental in many areas of science and engineering, from calculating the displacement of an object given its velocity to determining probabilities in statistical analysis. So, mastering these concepts is super crucial for anyone diving deep into these fields. The process of integration is essentially the reverse of differentiation, and understanding this relationship is key to solving these problems efficiently.

The function we are dealing with, 6xe^(x2), involves an exponential term with a composite function in the exponent. This is a clear signal that u-substitution will be our best friend in solving this integral. U-substitution allows us to simplify the integral by replacing a complex part of the function with a single variable, making the integration process much more manageable. It's like a clever trick that transforms a seemingly tough problem into something much simpler. Without u-substitution, this integral would be quite challenging to solve directly, so it’s a technique worth mastering. In the following sections, we’ll walk through exactly how to apply u-substitution to this specific problem, making sure each step is crystal clear.

Remember, integrals might seem daunting at first, but with practice and the right techniques, they become much more approachable. So, let's break this down together and make sure we've got a solid understanding of how to tackle these types of problems. By the end of this guide, you’ll be confidently evaluating definite integrals like this one, ready to impress your friends and ace your exams. Let's keep going!

Applying U-Substitution

The key to cracking this integral is a technique called u-substitution. It's like a mathematical magic trick that simplifies complex integrals. The first step is to identify a suitable part of the integrand (the function inside the integral) to substitute with a new variable, 'u'. In our case, the exponent x^2 is a prime candidate because its derivative, 2x, is also present in the integrand (we have 6x, which is just a constant multiple away). So, let’s set:

u = x^2

Now, we need to find the derivative of u with respect to x, which we denote as du/dx. Differentiating u = x^2 gives us:

du/dx = 2x

To isolate du, we multiply both sides by dx:

du = 2x dx

But wait! Our original integral has 6x dx, not 2x dx. No problem! We can easily adjust this by multiplying both sides of our equation by 3:

3 du = 6x dx

Now we have an expression that perfectly matches a part of our original integral. This is the magic of u-substitution in action! We’ve transformed a chunk of the integral into a simpler form. Next, we'll substitute u and 3 du into the original integral. This substitution will change the variable of integration from x to u, making the integral much easier to handle. The goal here is to rewrite the entire integral in terms of u, so there are no x variables lingering around. This makes the integration process straightforward.

Before we do that, though, there's one more crucial step. Since we're dealing with a definite integral, we need to change the limits of integration to match our new variable, u. Our original limits were in terms of x (0 and 2), but now we need them in terms of u. This is essential because we're changing the variable over which we're integrating, and the limits must correspond to the new variable. To find the new limits, we plug our original x limits into our substitution equation, u = x^2. When x = 0, u = 0^2 = 0. And when x = 2, u = 2^2 = 4. So, our new limits of integration are 0 and 4. This step ensures that we're calculating the area under the curve correctly in terms of the new variable.

With these transformations, our integral is about to look a whole lot simpler. We’ve successfully set up the u-substitution, found our du, and adjusted the limits of integration. Now we’re ready to rewrite the integral and actually solve it. This process demonstrates the power of u-substitution in simplifying integrals that would otherwise be quite tricky. Let's move on to the next step and see how this all comes together!

Rewriting and Evaluating the Integral

Alright, we've set the stage with u-substitution, and now it's time to rewrite our integral in terms of u. Remember, we had:

u = x^2
3 du = 6x dx
New limits: 0 to 4 (when x = 0, u = 0; when x = 2, u = 4)

Our original integral was:

∫[0 to 2] 6xe^(x2) dx

Now, let's substitute u and 3 du into the integral. We replace x^2 with u and 6x dx with 3 du. Our new integral looks like this:

∫[0 to 4] 3e^u du

See how much simpler that looks? We've transformed a relatively complex integral into a straightforward exponential integral. The constant 3 can be pulled out of the integral, which gives us:

3 ∫[0 to 4] e^u du

Now, we're ready to integrate. The integral of e^u with respect to u is simply e^u. So, we have:

3 [e^u] evaluated from 0 to 4

To evaluate the definite integral, we plug in the upper limit (4) and the lower limit (0) into our integrated function and subtract the results:

3 (e^4 - e^0)

Remember that e^0 is equal to 1. So, we have:

3 (e^4 - 1)

This is our exact answer. If we want a numerical approximation, we can use a calculator to find the value of e^4, which is approximately 54.598. Plugging this in, we get:

3 (54.598 - 1) = 3 (53.598) ≈ 160.794

So, the value of the definite integral ∫[0 to 2] 6xe^(x2) dx is approximately 160.794.

We've successfully evaluated the integral by using u-substitution to simplify the expression, changing the limits of integration, integrating, and then evaluating at the limits. This process highlights the power of u-substitution in making complex integrals manageable. The key steps were identifying the correct substitution, rewriting the integral, and remembering to change the limits of integration. By breaking down the problem into these steps, we made it much easier to solve. Congrats, you guys have nailed it!

Conclusion

Great job, you guys! We've successfully evaluated the definite integral ∫[0 to 2] 6xe^(x2) dx. By using the technique of u-substitution, we transformed a seemingly complex integral into something much more manageable. Let's quickly recap the key steps we took:

Identify the u-substitution: We recognized that x^2 was a good candidate for u because its derivative, 2x, was present in the integrand.
Find du: We calculated du/dx = 2x and then solved for du = 2x dx. We then adjusted this to match our integral, getting 3 du = 6x dx.
Change the limits of integration: We changed the limits from x = 0 and x = 2 to u = 0 and u = 4 by plugging the original limits into our substitution equation u = x^2.
Rewrite the integral: We substituted u and 3 du into the original integral, obtaining ∫[0 to 4] 3e^u du.
Evaluate the integral: We integrated 3e^u to get 3e^u, and then evaluated it at the new limits of integration, giving us 3(e^4 - e^0).
Simplify: We simplified the expression to get the exact answer, 3(e^4 - 1), and the approximate numerical value, 160.794.

This example showcases the power and elegance of u-substitution. It’s a technique that you'll use frequently in calculus, so it's super important to get comfortable with it. Remember, the key is to look for a part of the integrand whose derivative is also present (up to a constant multiple). Once you've identified a good candidate for u, the rest of the process usually falls into place quite smoothly.

Integrals might seem tricky at first, but with practice, they become much less daunting. By breaking down the problem into manageable steps and using clever techniques like u-substitution, you can tackle even the most intimidating integrals. Keep practicing, keep exploring, and you'll become a calculus whiz in no time! And remember, if you ever get stuck, there are tons of resources available to help you out, including textbooks, online tutorials, and your friendly neighborhood math enthusiasts (like us!). Happy integrating!