Definite Integrals: Solving With Given Values

Hey guys! Today, we're diving into some definite integral problems. We've got functions f(x) and g(x), and we know a few key integrals involving them. Our mission? To use these pieces of information to solve for some other integrals. Let's break it down step-by-step.

a. Evaluating $\int_4^4 f(x) dx$

In this first part, we're asked to find the definite integral of f(x) from 4 to 4: $\int_4^4 f(x) dx$. Now, remember a fundamental property of definite integrals: if the upper and lower limits of integration are the same, the value of the integral is always zero. Think of it like this: you're not actually traversing any distance along the x-axis, so there's no area under the curve to calculate. Therefore, the definite integral $\int_a^a f(x) dx = 0$ for any function f(x) that is defined at x = a. No matter how complex the function f(x) is, if the limits of integration are identical, the result will always be zero. This is because the integral represents the signed area under the curve, and if the interval has zero width (i.e., the upper and lower limits are the same), then the area is zero. This property holds true for all integrable functions, regardless of their specific form. So, whether f(x) is a simple polynomial, a trigonometric function, an exponential function, or a more complicated combination of these, the integral from 4 to 4 will always be zero. This is a direct consequence of the definition of the definite integral as the limit of a Riemann sum, where the width of each rectangle in the sum approaches zero as the number of rectangles approaches infinity. When the upper and lower limits are the same, the width of the interval is zero, and thus the entire sum becomes zero. This is a foundational concept in calculus and is essential for understanding the behavior of definite integrals. Thus, in our case:

$\int_4^4 f(x) dx = 0$

This is a direct application of the property of definite integrals, and it highlights the importance of understanding the fundamental concepts of calculus.

b. Evaluating $\int_2^7 2g(x) dx$

Next up, we need to evaluate $\int_2^7 2g(x) dx$. We are given that $\int_2^7 g(x) dx = 5$. Here, we can use another important property of definite integrals: the constant multiple rule. This rule states that for any constant c and integrable function g(x), the integral of c times g(x) is equal to c times the integral of g(x). Mathematically, this is expressed as: $\int_a^b c \cdot g(x) dx = c \cdot \int_a^b g(x) dx$. This property is incredibly useful because it allows us to pull constants out of integrals, simplifying the calculation process. In our specific problem, we have the integral of 2g(x) from 2 to 7. According to the constant multiple rule, we can rewrite this as 2 times the integral of g(x) from 2 to 7: $\int_2^7 2g(x) dx = 2 \cdot \int_2^7 g(x) dx$. We are given that $\int_2^7 g(x) dx = 5$, so we can substitute this value into our equation. This gives us: $2 \cdot \int_2^7 g(x) dx = 2 \cdot 5 = 10$. Therefore, the value of the integral $\int_2^7 2g(x) dx$ is 10. This demonstrates how the constant multiple rule can be applied to simplify definite integrals and make them easier to evaluate. By understanding and applying this rule, we can efficiently solve a wide range of integral problems. This property is a cornerstone of integral calculus and is essential for both theoretical understanding and practical applications.

So, we have:

$\int_2^7 2g(x) dx = 2 \int_2^7 g(x) dx = 2(5) = 10$

c. Evaluating $\int_4^7 f(x) dx$

Finally, let's tackle $\int_4^7 f(x) dx$. We know that $\int_2^4 f(x) dx = 3$ and $\int_2^7 f(x) dx = -7$. To find the integral from 4 to 7, we can use the property that states: $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$. In our case, we can rewrite the integral from 2 to 7 as the sum of the integral from 2 to 4 and the integral from 4 to 7: $\int_2^7 f(x) dx = \int_2^4 f(x) dx + \int_4^7 f(x) dx$. We want to find $\int_4^7 f(x) dx$, so we can rearrange the equation to solve for it: $\int_4^7 f(x) dx = \int_2^7 f(x) dx - \int_2^4 f(x) dx$. We are given that $\int_2^7 f(x) dx = -7$ and $\int_2^4 f(x) dx = 3$, so we can substitute these values into the equation: $\int_4^7 f(x) dx = -7 - 3 = -10$. Therefore, the value of the integral $\int_4^7 f(x) dx$ is -10. This demonstrates how we can use the properties of definite integrals to break down complex integrals into simpler ones and solve for unknown integrals using known values. This is a powerful technique in calculus and is essential for solving a wide range of problems. By understanding and applying these properties, we can efficiently evaluate definite integrals and gain a deeper understanding of the behavior of functions and their integrals. This technique is fundamental to many applications of calculus in physics, engineering, and other fields.

So here's the calculation:

$\int_4^7 f(x) dx = \int_2^7 f(x) dx - \int_2^4 f(x) dx = -7 - 3 = -10$

In summary:

• $\int_4^4 f(x) dx = 0$
• $\int_2^7 2g(x) dx = 10$
• $\int_4^7 f(x) dx = -10$

And that's it! By applying the properties of definite integrals, we were able to solve for the unknown integrals using the given information. Hope this helps, and happy integrating!