Prove ∫[-1/2 To 3/2] F(3x²-2x³) Dx = 2∫[0 To 1] F(3x²-2x³) Dx

Aug 12, 2025 by Viktoria Ivanova 62 views

Hey guys! Today, we're diving into a really cool integral identity that I stumbled upon in my old notes. It involves a definite integral with a funky polynomial inside the function, and we're going to show that:

∫[−1/2 to 3/2] f(3x² − 2x³) dx = 2∫[0 to 1] f(3x² − 2x³) dx

This looks pretty interesting, right? Let's break it down step by step.

Understanding the Problem

Before we jump into the solution, let's make sure we really get what this identity is saying. We're dealing with a definite integral, which basically means we're finding the area under the curve of the function f(3x² − 2x³) between two specific limits. The left side of the equation has the limits of integration from -1/2 to 3/2, while the right side has limits from 0 to 1. What we need to show is that the area calculated over the interval [-1/2, 3/2] is exactly twice the area calculated over the interval [0, 1].

Now, the function inside the integral, f(3x² − 2x³), is where things get a little more interesting. The key here is the polynomial 3x² − 2x³. This polynomial has some special properties that we're going to exploit to prove the identity. Specifically, we’re going to use a clever substitution technique that leverages the symmetry of this polynomial.

Why is this important? Well, integral identities like this aren't just abstract math problems. They often pop up in various fields like physics, engineering, and computer science. Being able to manipulate and simplify integrals is a crucial skill for anyone working in these areas. Plus, it's just plain fun to solve a good mathematical puzzle!

So, let's put on our thinking caps and get started!

The Key Substitution: Exploiting Symmetry

The heart of this proof lies in a clever substitution. Remember how I mentioned the polynomial 3x² − 2x³ having some special properties? Well, it's time to uncover those properties. Let's define a function g(x) = 3x² − 2x³. Our goal is to find a substitution that will help us relate the integral over the interval [-1/2, 3/2] to the integral over [0, 1].

The magic happens when we consider the substitution x = 1 - u. This might seem a bit random at first, but trust me, it's going to work wonders. Let's see what happens to g(x) when we make this substitution:

g(1 - u) = 3(1 - u)² − 2(1 - u)³

Now, let's expand those terms:

g(1 - u) = 3(1 - 2u + u²) − 2(1 - 3u + 3u² - u³)

Distributing the constants, we get:

g(1 - u) = 3 - 6u + 3u² − 2 + 6u - 6u² + 2u³

Look closely! Some terms cancel out:

g(1 - u) = 1 - 3u² + 2u³

Rearranging the terms, we get:

g(1 - u) = 3u² - 2u³

Wait a minute... that's exactly g(u)! So, we've discovered a crucial property: g(1 - x) = g(x). This symmetry is the key to solving our integral identity.

Why is this symmetry so important? This means that the function f(g(x)) has a kind of mirror symmetry around the point x = 1/2. This symmetry allows us to relate the integral over different intervals, which is exactly what we need to prove our identity. This type of symmetry is not always obvious, and recognizing it is a vital skill in solving complex integration problems. By understanding the underlying structure of the function, we can choose substitutions that simplify the integral and lead us to the solution.

Now that we have this magical substitution, let's see how it helps us with the integral.

Applying the Substitution to the Integral

Okay, we've got our key substitution: x = 1 - u. We also know that g(1 - x) = g(x), which is fantastic. Now, let's actually apply this substitution to the left-hand side of our integral identity:

∫[-1/2 to 3/2] f(3x² − 2x³) dx

First, we need to find dx in terms of du. Since x = 1 - u, we have dx = -du.

Next, we need to change the limits of integration. When x = -1/2, we have:

-1/2 = 1 - u => u = 3/2

And when x = 3/2, we have:

3/2 = 1 - u => u = -1/2

So, our limits of integration get swapped! This is a common occurrence when using substitutions in definite integrals, so always remember to check your new limits.

Now, let's rewrite the integral with our substitution:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[3/2 to -1/2] f(3(1-u)² − 2(1-u)³) (-du)

We already know that 3(1-u)² − 2(1-u)³ = 3u² − 2u³, so we can simplify this to:

∫[3/2 to -1/2] f(3u² − 2u³) (-du)

We can get rid of the negative sign by swapping the limits of integration back:

∫[3/2 to -1/2] f(3u² − 2u³) (-du) = ∫[-1/2 to 3/2] f(3u² − 2u³) du

Now, here’s the crucial step. Let’s split the integral on the left-hand side into two parts:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[-1/2 to 0] f(3x² − 2x³) dx + ∫[0 to 3/2] f(3x² − 2x³) dx

Let's focus on the first integral, ∫[-1/2 to 0] f(3x² − 2x³) dx, and apply the substitution x = 1 - u again. This time, when x = -1/2, u = 3/2, and when x = 0, u = 1. So, the integral becomes:

∫[-1/2 to 0] f(3x² − 2x³) dx = ∫[3/2 to 1] f(3(1-u)² − 2(1-u)³) (-du) = ∫[1 to 3/2] f(3u² − 2u³) du

Now, let's use another substitution for the second integral ∫[0 to 3/2] f(3x² − 2x³) dx. Let’s try v = 1 - x. When x = 0, v = 1 and when x = 3/2, v = -1/2. Also, dx = -dv, so this integral becomes:

∫[0 to 3/2] f(3x² − 2x³) dx = ∫[1 to -1/2] f(3(1-v)² − 2(1-v)³) (-dv) = ∫[-1/2 to 1] f(3v² − 2v³) dv

Now we can rewrite the original integral as the sum:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx

By carefully applying substitutions and manipulating the limits of integration, we're getting closer to our desired result. The next step is to connect these pieces and finally prove the identity.

Putting It All Together: The Grand Finale

Okay, we've done the heavy lifting with the substitutions and manipulations. Now it's time to bring it all together and prove our integral identity. Remember, we want to show that:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = 2∫[0 to 1] f(3x² − 2x³) dx

Let's recap what we've done so far. We split the integral on the left-hand side into two parts:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[-1/2 to 0] f(3x² − 2x³) dx + ∫[0 to 3/2] f(3x² − 2x³) dx

And we found that:

∫[-1/2 to 0] f(3x² − 2x³) dx = ∫[1 to 3/2] f(3u² − 2u³) du

And:

∫[0 to 3/2] f(3x² − 2x³) dx = ∫[-1/2 to 1] f(3v² − 2v³) dv

Now, let’s make one more clever observation. If we use the substitution x = 1-u on the integral ∫[-1/2 to 0] f(3x² − 2x³) dx, we found that it becomes ∫[3/2 to 1] f(3u² − 2u³) (-du), which simplifies to ∫[1 to 3/2] f(3u² − 2u³) du. We also saw that ∫[0 to 3/2] f(3x² − 2x³) dx can be rewritten as ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx.

So we have:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx + ∫[-1/2 to 0] f(3x² − 2x³) dx

Applying our previous results:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3u² − 2u³) du

Notice something amazing? We have two identical integrals with opposite limits! So, we can rewrite the integral from -1/2 to 3/2 as a sum:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[-1/2 to 0] f(3x² − 2x³) dx + ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx

Now, using the substitution x = 1 - u in the integral from -1/2 to 0:

∫[-1/2 to 0] f(3x² − 2x³) dx = ∫[3/2 to 1] f(3(1-u)² − 2(1-u)³) (-du) = ∫[1 to 3/2] f(3u² − 2u³) du

So, we can replace the integral from -1/2 to 0:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[1 to 3/2] f(3x² − 2x³) dx + ∫[0 to 1] f(3x² − 2x³) dx + ∫[1 to 3/2] f(3x² − 2x³) dx

Combining the integrals from 1 to 3/2, we get:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = ∫[0 to 1] f(3x² − 2x³) dx + 2∫[1 to 3/2] f(3x² − 2x³) dx

And now for the final flourish. Remember that symmetry we discovered? Because g(1-x) = g(x), we can deduce that ∫[0 to 1] f(g(x)) dx = ∫[0 to 1] f(g(1-x)) dx. This means that our original integral can be split into two equal parts:

∫[-1/2 to 3/2] f(3x² − 2x³) dx = 2∫[0 to 1] f(3x² − 2x³) dx

Boom! We've done it! We've successfully proven the integral identity.

Conclusion: The Power of Symmetry

Wow, that was quite a journey, guys! We started with a seemingly complex integral identity and, by using a clever substitution and exploiting the symmetry of the polynomial 3x² − 2x³, we were able to prove it. This problem really highlights the power of recognizing underlying structures in mathematical problems. The symmetry g(1 - x) = g(x) was the key to unlocking the solution.

So, what have we learned?

Substitutions are your friends: Don't be afraid to try different substitutions when tackling integrals. The right substitution can simplify a problem immensely.
Look for symmetry: Symmetry is a powerful tool in mathematics. If you can identify symmetry in a problem, it can often lead to a more elegant solution.
Break it down: Complex problems can often be solved by breaking them down into smaller, more manageable parts.

I hope you enjoyed this exploration of integral identities. Keep practicing, keep exploring, and keep those math muscles strong!