Evaluating The Definite Integral Of 3/(1+16x^2) From 0 To 1/4

Hey guys! Today, we're diving into a fun little calculus problem: evaluating a definite integral. Specifically, we're going to tackle the integral of 3 divided by (1 plus 16 times x squared), from 0 to 1/4. Buckle up, because we're about to embark on a mathematical journey filled with u-substitutions, arctangent functions, and the sheer joy of problem-solving!

Understanding the Integral

Let's break down the definite integral we're working with: ∫[0 to 1/4] 3/(1+16x²) dx. The key to solving this integral lies in recognizing its resemblance to the derivative of the arctangent function. Remember, the derivative of arctan(u) with respect to x is du/dx divided by (1 + u²). Our integral has a similar form, which suggests we can use a substitution to make it look exactly like the arctangent derivative. The arctan function is the inverse tangent function, also written as tan⁻¹(x). It gives you the angle whose tangent is x. So, when we see an integral that looks like something over (1 + something squared), our minds should immediately jump to arctan!. The definite integral means we're finding the area under the curve of the function 3/(1+16x²) between the limits of integration, which are 0 and 1/4 in this case.

The U-Substitution Technique

The first trick that we will use is to apply u-substitution, this technique is super handy for integrals that seem complicated at first glance. The main idea here is to simplify our integral by replacing a part of the expression with a new variable, which we'll call 'u'. In this particular problem, the term '16x²' in the denominator is a bit of a giveaway. It suggests that we can make a substitution to bring our integral closer to the familiar form of the arctangent integral. Let's try setting u equal to 4x. This is a strategic choice because when we square 'u', we get 16x², which matches the term in our denominator. Now, we need to find the relationship between 'du' and 'dx'. If u = 4x, then taking the derivative of both sides with respect to x gives us du/dx = 4. This means that du = 4 dx. But our original integral has 'dx', not '4 dx'. No worries! We can easily adjust this by dividing both sides by 4, giving us dx = (1/4) du. Okay, we've got our 'u' and our 'dx' in terms of 'du'. But there's one more crucial step when dealing with definite integrals and u-substitution: we need to change the limits of integration. Our original limits were in terms of 'x', but now that we're integrating with respect to 'u', we need the corresponding 'u' values. When x = 0 (our lower limit), u = 4 * 0 = 0. So the new lower limit is still 0. When x = 1/4 (our upper limit), u = 4 * (1/4) = 1. So the new upper limit is 1. Now we have everything we need to rewrite the integral in terms of 'u'. This substitution is a powerful tool because it transforms a complex integral into a simpler, more recognizable form. By carefully choosing our 'u' and adjusting the limits of integration, we've set ourselves up for a much easier integration process.

Rewriting and Simplifying the Integral

Now, let's plug in our u-substitution into the integral. Remember, we have u = 4x, 16x² = u², and dx = (1/4) du. Replacing these into our original integral ∫[0 to 1/4] 3/(1+16x²) dx, we get: ∫[0 to 1] 3/(1+u²) * (1/4) du. See how much cleaner that looks already? The power of u-substitution is on full display here. We've transformed a slightly intimidating integral into a much more manageable form. But we're not quite done yet. We can simplify this even further by pulling out the constants. Notice that we have a '3' in the numerator and a '1/4' from our dx substitution. Both of these are constants with respect to 'u', so we can move them outside the integral. This gives us: (3/4) ∫[0 to 1] 1/(1+u²) du. Ah, that's much better! Now we're left with a constant multiplied by a very familiar integral. This simplified form makes it much easier to recognize the integral as related to the arctangent function. By carefully substituting and simplifying, we've made the integral far less intimidating and brought it into a form that we can directly integrate. This is a common strategy in calculus – break down complex problems into simpler, more recognizable pieces.

Integrating with Arctangent

Alright, guys, here's where the magic happens! We've transformed our integral into (3/4) ∫[0 to 1] 1/(1+u²) du. Now, let's think about what function has a derivative of 1/(1+u²). Ding ding ding! It's the arctangent function, or tan⁻¹(u). This is a crucial step – recognizing the integral form and knowing its corresponding antiderivative. Remember, the integral of 1/(1+x²) dx is arctan(x) + C, where C is the constant of integration. In our case, we have the integral of 1/(1+u²) du, so the antiderivative is simply arctan(u). Now we can apply the Fundamental Theorem of Calculus, which tells us that to evaluate a definite integral, we need to find the antiderivative of the function and then evaluate it at the upper and lower limits of integration. So, the integral of (3/4) * [1/(1+u²)] du is (3/4) * arctan(u). We don't need to worry about the constant of integration 'C' because it will cancel out when we evaluate the definite integral. Now we need to evaluate (3/4) * arctan(u) at our limits of integration, which are 0 and 1. This means we need to find (3/4) * arctan(1) and (3/4) * arctan(0), and then subtract the value at the lower limit from the value at the upper limit. By identifying the antiderivative and applying the Fundamental Theorem of Calculus, we're on the home stretch to solving this integral. This step highlights the importance of knowing your common derivatives and antiderivatives – they're the tools of the trade in calculus!

Evaluating the Limits and Final Answer

Okay, time to plug in those limits of integration! We've got (3/4) * arctan(u) evaluated from 0 to 1. First, let's find arctan(1). Remember, arctan(x) asks the question: