Definite Integrals And Inverse Functions Evaluating ∫f(x)dx + ∫f⁻¹(x)dx

Hey guys! Let's dive into a fascinating problem involving definite integrals and inverse functions. This topic often appears in calculus, and mastering it can significantly boost your problem-solving skills. In this article, we'll break down the problem step-by-step, making sure you understand every nuance. So, buckle up and let's get started!

The Core Problem

The heart of our discussion lies in evaluating a specific expression involving both a function and its inverse. We're given the function $f(x) = x^3 + 3x + 4$, and our mission, should we choose to accept it, is to find the value of the following expression:

$$\int_{-1}^{1} f(x) dx + \int_{0}^{4} f^{-1}(x) dx$$

This looks a bit intimidating at first glance, right? We have the integral of $f(x)$ over a certain interval and the integral of its inverse, $f^{-1}(x)$, over another interval. The key here is to recognize that these integrals are related geometrically. Visualizing what's going on can make the solution click much faster. We'll get into the nitty-gritty details soon, but first, let’s set the stage.

Understanding the Function

Before we even think about integrals, let's get to know our function, $f(x) = x^3 + 3x + 4$. This is a cubic function, and cubic functions have some interesting properties. Notice that the derivative of $f(x)$ is:

$$f'(x) = 3x^2 + 3$$

Since $3x^2$ is always non-negative, $f'(x)$ is always positive. This tells us that $f(x)$ is a strictly increasing function. What does that mean? Simply put, as $x$ increases, $f(x)$ also increases. This is crucial because strictly increasing functions have inverses! The inverse function, denoted as $f^{-1}(x)$, essentially "undoes" what $f(x)$ does. If $f(a) = b$, then $f^{-1}(b) = a$.

The Geometric Connection

Now, let's bring in the geometry. The definite integral $\int_{a}^{b} f(x) dx$ represents the area under the curve of $f(x)$ between the vertical lines $x = a$ and $x = b$. Similarly, the integral $\int_{c}^{d} f^{-1}(x) dx$ represents the area under the curve of $f^{-1}(x)$ between the vertical lines $x = c$ and $x = d$. But how are these areas related?

The magic happens when we realize that the graph of $f^{-1}(x)$ is a reflection of the graph of $f(x)$ across the line $y = x$. This reflection property is fundamental to understanding inverse functions. Imagine you have a point $(a, b)$ on the graph of $f(x)$. This means $f(a) = b$. The corresponding point on the graph of $f^{-1}(x)$ is $(b, a)$, which you get by swapping the $x$ and $y$ coordinates. This swap is exactly what a reflection across the line $y = x$ does!

So, the areas under the curves of $f(x)$ and $f^{-1}(x)$ are intricately linked. This link is what we’ll exploit to solve our problem. We’re not just dealing with two separate integrals; we’re dealing with two integrals that represent areas of related shapes.

Solving the Integral Problem

Okay, now let's get down to business and actually calculate the value of the expression.

$$\int_{-1}^{1} f(x) dx + \int_{0}^{4} f^{-1}(x) dx$$

Breaking Down the Integrals

First, let's consider the integral $\int_{-1}^{1} f(x) dx$. We have $f(x) = x^3 + 3x + 4$, so this integral becomes:

$$\int_{-1}^{1} (x^3 + 3x + 4) dx$$

We can break this up into simpler integrals:

$$\int_{-1}^{1} x^3 dx + \int_{-1}^{1} 3x dx + \int_{-1}^{1} 4 dx$$

Let's evaluate each part separately. The integral of $x^3$ is $\frac{x^4}{4}$. The integral of $3x$ is $\frac{3x^2}{2}$. And the integral of $4$ is simply $4x$. So we have:

$$\left[\frac{x^4}{4}\right]_{-1}^{1} + \left[\frac{3x^2}{2}\right]_{-1}^{1} + \left[4x\right]_{-1}^{1}$$

Now, we plug in the limits of integration:

$$\left(\frac{1^4}{4} - \frac{(-1)^4}{4}\right) + \left(\frac{3(1)^2}{2} - \frac{3(-1)^2}{2}\right) + \left(4(1) - 4(-1)\right)$$

Simplifying this, we get:

$$( \frac{1}{4} - \frac{1}{4} ) + ( \frac{3}{2} - \frac{3}{2} ) + (4 + 4) = 0 + 0 + 8 = 8$$

So, $\int_{-1}^{1} f(x) dx = 8$.

Tackling the Inverse Function Integral

Now for the trickier part: $\int_{0}^{4} f^{-1}(x) dx$. Directly integrating an inverse function can be a real headache. But remember our geometric connection? This is where it shines.

The formula we'll use is a gem in calculus: If $f(a) = c$ and $f(b) = d$, then:

$$\int_{a}^{b} f(x) dx + \int_{c}^{d} f^{-1}(x) dx = bf(b) - af(a) = bd - ac$$

This formula beautifully connects the integrals of a function and its inverse. It comes from looking at the areas and rectangles formed by the function, its inverse, and the coordinate axes. Trust me, it's a powerful tool!

To use this formula, we need to find values of $a$, $b$, $c$, and $d$ that fit our integral $\int_{0}^{4} f^{-1}(x) dx$. We know that the limits of integration for the inverse function are $0$ and $4$. So, $c = 0$ and $d = 4$. We need to find the corresponding $a$ and $b$ such that $f(a) = 0$ and $f(b) = 4$.

Let’s solve $f(x) = 0$:

$$x^3 + 3x + 4 = 0$$

By observation or using the rational root theorem, we find that $x = -1$ is a solution. So, $f(-1) = 0$, which means $a = -1$.

Next, let's solve $f(x) = 4$:

$$x^3 + 3x + 4 = 4$$

$$x^3 + 3x = 0$$

$$x(x^2 + 3) = 0$$

The solutions are $x = 0$ (since $x^2 + 3$ is always positive). So, $f(0) = 4$, which means $b = 0$.

Now we have all our values: $a = -1$, $b = 0$, $c = 0$, and $d = 4$. Plugging these into our formula, we get:

$$\int_{-1}^{0} f(x) dx + \int_{0}^{4} f^{-1}(x) dx = (0)(4) - (-1)(0) = 0$$

Wait a minute! We don't need the first integral on the left side, we only need $\int_{0}^{4} f^{-1}(x) dx$! Let's rewrite the equation we are using:

$$\int_{c}^{d} f^{-1}(x) dx = bd - ac - \int_{a}^{b} f(x) dx$$

Substituting $a=-1$ and $b=0$ into $f(x)$:

$$\int_{-1}^{0} (x^3 + 3x + 4) dx = \left[\frac{x^4}{4} + \frac{3x^2}{2} + 4x\right]_{-1}^{0}$$

Evaluating this gives:

$$(0) - (\frac{1}{4} + \frac{3}{2} - 4) = -(\frac{1}{4} + \frac{6}{4} - \frac{16}{4}) = -(\frac{-9}{4}) = \frac{9}{4}$$

Therefore, $\int_{-1}^{0} f(x) dx = \frac{9}{4}$.

Now, we can use this result to find $\int_{0}^{4} f^{-1}(x) dx$. Using the formula:

$$\int_{0}^{4} f^{-1}(x) dx = (0)(4) - (-1)(0) - \frac{9}{4} = -\frac{9}{4}$$

Putting It All Together

Finally, we can find the value of the original expression:

$$\int_{-1}^{1} f(x) dx + \int_{0}^{4} f^{-1}(x) dx = 8 + (-\frac{9}{4}) = \frac{32}{4} - \frac{9}{4} = \frac{23}{4}$$

So, the value of the expression is $\frac{23}{4}$.

Key Takeaways

That was quite a journey, wasn't it? Let's recap the key concepts we've explored:

1. Inverse Functions: Understanding that the inverse function "undoes" the original function and the geometric reflection property is crucial.
2. Definite Integrals as Areas: Visualizing definite integrals as areas under curves helps connect the integrals of $f(x)$ and $f^{-1}(x)$.
3. The Magic Formula: The formula $\int_{a}^{b} f(x) dx + \int_{c}^{d} f^{-1}(x) dx = bd - ac$ (where $f(a) = c$ and $f(b) = d$) is a powerful shortcut for evaluating integrals involving inverse functions.
4. Step-by-Step Approach: Breaking down complex problems into smaller, manageable steps makes the solution clearer and less daunting.

An Additional Integral Problem

Let's look at another integral problem to further solidify our understanding. Consider the integral:

$$\int_{0}^{1} (e-1) \sqrt{\ln[1 + (e-1)x]} dx$$

This integral looks quite different from our first one, but the same principles of careful observation and strategic substitution can help us solve it. Let's break it down:

1. Recognizing Complexity: The presence of the nested function $\sqrt{\ln[1 + (e-1)x]}$ immediately suggests that a direct approach might be cumbersome. We need to simplify this expression.

2. Strategic Substitution: The key here is to make a substitution that simplifies the innermost function. A common technique is to let $u$ equal the most complex part. So, let's try:

$$u = \ln[1 + (e-1)x]$$

3. Finding dx: We need to express $dx$ in terms of $du$. First, we rewrite the equation:

$$e^u = 1 + (e-1)x$$

Now, solve for $x$:

$$x = \frac{e^u - 1}{e - 1}$$

4. Differentiate: Differentiate both sides with respect to $u$:

$$\frac{dx}{du} = \frac{e^u}{e - 1}$$

So, $dx = \frac{e^u}{e - 1} du$.

5. Change Limits of Integration: When we substitute, we also need to change the limits of integration. When $x = 0$:

$$u = \ln[1 + (e-1)(0)] = \ln(1) = 0$$

When $x = 1$:

$$u = \ln[1 + (e-1)(1)] = \ln(e) = 1$$

So, our new limits of integration are $0$ and $1$.

6. Rewrite the Integral: Now we can rewrite the integral in terms of $u$:

$$\int_{0}^{1} (e-1) \sqrt{u} \frac{e^u}{e - 1} du$$

Notice that the $(e-1)$ terms cancel out, simplifying the integral:

$$\int_{0}^{1} \sqrt{u} e^u du$$

7. Integration by Parts: This integral looks more manageable, but we still need to tackle the product of $\sqrt{u}$ and $e^u$. Integration by parts is the perfect tool here. The formula for integration by parts is:

$$\int v dw = vw - \int w dv$$

Let's choose $v = \sqrt{u}$ and $dw = e^u du$. Then:

$$dv = \frac{1}{2\sqrt{u}} du$$

$$w = \int e^u du = e^u$$

8. Apply Integration by Parts: Plugging these into the formula, we get:

$$\int_{0}^{1} \sqrt{u} e^u du = \left[\sqrt{u} e^u\right]_{0}^{1} - \int_{0}^{1} e^u \frac{1}{2\sqrt{u}} du$$

Evaluating the first term:

$$\left[\sqrt{u} e^u\right]_{0}^{1} = \sqrt{1} e^1 - \sqrt{0} e^0 = e$$

Now, we have:

$$e - \frac{1}{2} \int_{0}^{1} \frac{e^u}{\sqrt{u}} du$$

9. Another Substitution (Optional): The remaining integral, $\int_{0}^{1} \frac{e^u}{\sqrt{u}} du$, is still a bit tricky, but we've made significant progress. While it doesn't have a simple closed-form solution in terms of elementary functions, we can express it in terms of special functions like the Error Function or approximate it numerically.

10. Final Result: For the sake of this explanation, we'll leave the final integral as is. Thus, the solution for the integral is:

$$\int_{0}^{1} (e-1) \sqrt{\ln[1 + (e-1)x]} dx = e - \frac{1}{2} \int_{0}^{1} \frac{e^u}{\sqrt{u}} du$$

Concluding Thoughts

Guys, we've covered a lot of ground! We started with a challenging integral problem involving inverse functions, used a clever geometric relationship and a powerful formula to solve it, and then tackled another complex integral using substitution and integration by parts. These techniques are fundamental in calculus, and mastering them will open doors to solving a wide range of problems.

Remember, the key is to break down problems into smaller steps, use strategic substitutions, and visualize the underlying concepts. Keep practicing, and you'll become a calculus whiz in no time! Keep up the great work and always be curious!