Finding Integration Bounds: A Calculus Problem Solved

by Alex Johnson 54 views

Let's dive into an intriguing problem about finding integration bounds. This is a classic calculus question that tests your understanding of definite integrals and their properties. We'll break down the problem step by step, making sure you grasp the core concepts and can tackle similar questions with confidence.

Understanding the Problem

The problem presents us with an equation involving definite integrals:

∫abf(x)dx=∫−36f(x)dx+∫67f(x)dx−∫−3−1f(x)dx\int_a^b f(x) dx = \int_{-3}^6 f(x) dx + \int_6^7 f(x) dx - \int_{-3}^{-1} f(x) dx

Our mission is to determine the values of 'a' and 'b', which represent the lower and upper bounds of integration for the integral on the left-hand side. To achieve this, we'll need to manipulate the integrals on the right-hand side, leveraging the fundamental properties of definite integrals. This will involve reversing limits, combining integrals, and a bit of algebraic thinking. So, buckle up, and let's get started!

Breaking Down Definite Integrals

Before we jump into solving the problem, let's quickly recap what definite integrals represent. A definite integral, denoted as ∫abf(x)dx{\int_a^b f(x) dx}, calculates the signed area between the curve of the function f(x) and the x-axis, from x = a to x = b. The values 'a' and 'b' are the limits of integration, defining the interval over which we're calculating the area. Understanding this geometric interpretation is crucial for grasping the properties we'll be using.

Key Properties of Definite Integrals

Several key properties of definite integrals will be instrumental in solving our problem. Let's highlight the most relevant ones:

  1. Reversing Limits: ∫abf(x)dx=−∫baf(x)dx{\int_a^b f(x) dx = -\int_b^a f(x) dx}. This property tells us that if we swap the limits of integration, we change the sign of the integral. This might seem simple, but it's incredibly useful for rearranging terms and combining integrals.
  2. Additivity: ∫abf(x)dx+∫bcf(x)dx=∫acf(x)dx{\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx}. This property states that if we have two integrals where the upper limit of the first integral matches the lower limit of the second integral, we can combine them into a single integral with the limits from the first integral's lower limit to the second integral's upper limit. This is like adding areas together seamlessly.
  3. Constant Multiple: ∫abk⋅f(x)dx=k∫abf(x)dx{\int_a^b k \cdot f(x) dx = k \int_a^b f(x) dx}, where k is a constant. This property allows us to pull a constant factor out of the integral, simplifying calculations.

With these properties in our arsenal, we're well-equipped to tackle the problem at hand. Remember, the key is to strategically apply these rules to manipulate the given equation and isolate the desired integration bounds.

Solving for the Integration Bounds

Now, let's get our hands dirty and solve for the integration bounds 'a' and 'b'. We'll start by carefully examining the given equation:

∫abf(x)dx=∫−36f(x)dx+∫67f(x)dx−∫−3−1f(x)dx\int_a^b f(x) dx = \int_{-3}^6 f(x) dx + \int_6^7 f(x) dx - \int_{-3}^{-1} f(x) dx

The first thing we notice is the subtraction of an integral. To make things easier to combine, let's use the "Reversing Limits" property to change the sign and flip the limits of integration in the last term:

∫abf(x)dx=∫−36f(x)dx+∫67f(x)dx+∫−1−3f(x)dx\int_a^b f(x) dx = \int_{-3}^6 f(x) dx + \int_6^7 f(x) dx + \int_{-1}^{-3} f(x) dx

Now, we have a sum of integrals. This looks promising! Let's try to use the "Additivity" property to combine them. Notice that we have an integral from -3 to 6 and another from 6 to 7. These can be combined:

∫−36f(x)dx+∫67f(x)dx=∫−37f(x)dx\int_{-3}^6 f(x) dx + \int_6^7 f(x) dx = \int_{-3}^7 f(x) dx

So, our equation now looks like this:

∫abf(x)dx=∫−37f(x)dx+∫−1−3f(x)dx\int_a^b f(x) dx = \int_{-3}^7 f(x) dx + \int_{-1}^{-3} f(x) dx

We're getting closer! Now, let's focus on the remaining integrals on the right-hand side. We have an integral from -3 to 7 and another from -1 to -3. Notice that the upper limit of the second integral (-3) matches the lower limit of the first integral. This is perfect for applying the "Additivity" property again, but we need to swap the order of the integrals first. Remember, the order matters when applying the additivity property. Let's rewrite the equation:

$\int_a^b f(x) dx = \int_{-1}^{-3} f(x) dx + \int_{-3}^7 f(x) dx $

Now, we can combine these integrals:

∫−1−3f(x)dx+∫−37f(x)dx=∫−17f(x)dx\int_{-1}^{-3} f(x) dx + \int_{-3}^7 f(x) dx = \int_{-1}^7 f(x) dx

Our equation simplifies beautifully to:

∫abf(x)dx=∫−17f(x)dx\int_a^b f(x) dx = \int_{-1}^7 f(x) dx

Aha! We've arrived at a point where we can directly compare the two integrals. For the equation to hold true, the limits of integration must be equal. Therefore:

a = -1 b = 7

And there you have it! We've successfully determined the integration bounds for the first integral.

Visualizing the Solution

It's often helpful to visualize what we've just done. Imagine the area under the curve f(x). We started with several different integrals representing areas over different intervals. By using the properties of definite integrals, we effectively rearranged and combined these areas until we had a single integral with the same bounds as the integral we were trying to solve for. This visual representation can solidify your understanding of the process.

Key Takeaways

Let's recap the key takeaways from this problem:

  • Understanding Definite Integrals: Remember that definite integrals represent the signed area under a curve between two limits.
  • Properties are Your Friends: The properties of definite integrals (Reversing Limits, Additivity, Constant Multiple) are powerful tools for manipulating and simplifying integral expressions.
  • Strategic Application: The key to solving these types of problems is to strategically apply the properties in a way that allows you to combine or rearrange integrals.
  • Visualize When Possible: Visualizing the areas represented by the integrals can help you understand the steps you're taking.

By mastering these concepts, you'll be well-prepared to tackle a wide range of calculus problems involving definite integrals and integration bounds. Keep practicing, and you'll become a pro in no time!

Practice Problems

To solidify your understanding, try tackling similar problems. For example, consider these variations:

  1. If ∫cdg(x)dx=∫25g(x)dx−∫23g(x)dx+∫73g(x)dx\int_c^d g(x) dx = \int_2^5 g(x) dx - \int_2^3 g(x) dx + \int_7^3 g(x) dx, find the values of 'c' and 'd'.
  2. Given ∫pqh(x)dx=∫−40h(x)dx+∫02h(x)dx−∫−4−1h(x)dx\int_p^q h(x) dx = \int_{-4}^0 h(x) dx + \int_0^2 h(x) dx - \int_{-4}^{-1} h(x) dx, determine the bounds 'p' and 'q'.

Working through these practice problems will reinforce your skills and build your confidence in solving integral-related questions.

Conclusion

Finding integration bounds involves a solid grasp of definite integral properties and strategic manipulation. By understanding the concepts and practicing regularly, you can confidently solve these types of problems. Remember to visualize the process and leverage the power of the integral properties. Happy integrating!

For further exploration and practice on definite integrals, you can visit resources like Khan Academy's Calculus section. They offer comprehensive lessons and exercises to enhance your understanding of calculus concepts.