Evaluating Integrals A Comprehensive Step-by-Step Guide

by Sharif Sakr 56 views

Hey guys! Today, we're diving deep into the fascinating world of integral calculus. We're going to tackle some tricky integrals, breaking them down step-by-step so you can conquer them too. Whether you're a student battling calculus or just a math enthusiast, this guide is for you. We'll be focusing on three specific integrals, each requiring different techniques and strategies. So, grab your notebooks, and let's get started!

1. Evaluating the Integral of ∫x³e²ˣ dx

Alright, let's kick things off with our first integral: ∫x³e²ˣ dx. This one looks a bit intimidating, doesn't it? You see a polynomial (x³) multiplied by an exponential function (e²ˣ). When you encounter such a mix, the golden ticket is often integration by parts. This technique allows us to shift the complexity from one part of the integral to another, hopefully making it easier to solve.

Understanding Integration by Parts

Before we jump into the nitty-gritty, let's quickly recap the integration by parts formula. It's derived from the product rule of differentiation and looks like this:

∫u dv = uv - ∫v du

Where:

  • u is a function we choose to differentiate.
  • dv is the remaining part of the integral, which we'll integrate.
  • du is the derivative of u.
  • v is the integral of dv.

The key here is choosing the right 'u' and 'dv'. The goal is to pick a 'u' that becomes simpler when differentiated. In our case, x³ seems like a good candidate because each time we differentiate it, the power decreases.

Applying Integration by Parts Iteratively

Let's apply this to our integral ∫x³e²ˣ dx. We'll set:

  • u = x³
  • dv = e²ˣ dx

Now, we need to find du and v:

  • du = 3x² dx
  • v = ∫e²ˣ dx = (1/2)e²ˣ

Plug these into the integration by parts formula:

∫x³e²ˣ dx = x³(1/2)e²ˣ - ∫(1/2)e²ˣ(3x²) dx

Simplifying, we get:

∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/2)∫x²e²ˣ dx

Notice that we still have an integral, ∫x²e²ˣ dx, but the power of x has decreased. This is progress! But we're not done yet. We need to apply integration by parts again.

Let's set:

  • u = x²
  • dv = e²ˣ dx

Then:

  • du = 2x dx
  • v = (1/2)e²ˣ

Applying integration by parts again:

∫x²e²ˣ dx = x²(1/2)e²ˣ - ∫(1/2)e²ˣ(2x) dx

∫x²e²ˣ dx = (1/2)x²e²ˣ - ∫xe²ˣ dx

Now, substitute this back into our original equation:

∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/2)[(1/2)x²e²ˣ - ∫xe²ˣ dx]

∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/2)∫xe²ˣ dx

Guess what? We need to do integration by parts one more time! This is tedious, but we're getting there.

Let's set:

  • u = x
  • dv = e²ˣ dx

Then:

  • du = dx
  • v = (1/2)e²ˣ

Applying integration by parts:

∫xe²ˣ dx = x(1/2)e²ˣ - ∫(1/2)e²ˣ dx

∫xe²ˣ dx = (1/2)xe²ˣ - (1/4)e²ˣ

Finally, substitute this back into our equation:

∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/2)[(1/2)xe²ˣ - (1/4)e²ˣ] + C

Simplify to get our final answer:

∫x³e²ˣ dx = (1/2)x³e²ˣ - (3/4)x²e²ˣ + (3/4)xe²ˣ - (3/8)e²ˣ + C

There you have it! We conquered that beast of an integral using integration by parts repeatedly. Remember, the key is to choose 'u' wisely and be patient with the process.

2. Evaluating the Definite Integral ∫₀¹ (2x)/(x²+1)² dx

Now, let's switch gears and tackle a definite integral: ∫₀¹ (2x)/(x²+1)² dx. This one looks different from the first one, right? We have a fraction with a squared term in the denominator. When you see this kind of structure, u-substitution often comes to the rescue.

Mastering U-Substitution

U-substitution is a powerful technique that allows us to simplify integrals by changing the variable of integration. The idea is to identify a part of the integrand (the function being integrated) whose derivative also appears in the integral. This allows us to rewrite the integral in terms of a new variable, 'u', making it easier to solve.

Applying U-Substitution

In our integral ∫₀¹ (2x)/(x²+1)² dx, notice that the derivative of x² + 1 is 2x, which is conveniently present in the numerator. This is a clear signal for u-substitution!

Let's set:

  • u = x² + 1

Then:

  • du = 2x dx

See how the 2x dx in the numerator perfectly matches our du? This is what we want!

Now, we need to change the limits of integration. Since we're switching from x to u, we need to find the corresponding u values for our x limits:

  • When x = 0, u = 0² + 1 = 1
  • When x = 1, u = 1² + 1 = 2

So, our new integral in terms of u becomes:

∫₁² (1/u²) du

This looks much simpler, doesn't it? We can rewrite 1/u² as u⁻² and apply the power rule for integration:

∫₁² u⁻² du = [-u⁻¹]₁²

Now, we evaluate the antiderivative at the upper and lower limits and subtract:

[-u⁻¹]₁² = [-1/u]₁² = (-1/2) - (-1/1) = -1/2 + 1 = 1/2

Therefore, the value of the definite integral ∫₀¹ (2x)/(x²+1)² dx is 1/2. We successfully used u-substitution to transform a seemingly complex integral into a straightforward one.

3. Finding the Area Under a Curve y = h(x)

Our final challenge is a bit more conceptual. We're asked to find the area under the curve y = h(x) from... well, we don't have specific limits! This is a deliberate move. The point here is to understand the fundamental concept of how integrals relate to areas.

The Definite Integral as Area

The definite integral of a function between two points represents the signed area between the function's graph and the x-axis.