Problem 49

Question

Find \(d y / d x\). $$y=\int_{-1}^{x} \frac{t^{2}}{t^{2}+4} d t-\int_{3}^{x} \frac{t^{2}}{t^{2}+4} d t$$

Step-by-Step Solution

Verified
Answer
The derivative is 0.
1Step 1: Understanding the Problem
We need to differentiate the function \( y = \int_{-1}^{x} \frac{t^{2}}{t^{2}+4} dt - \int_{3}^{x} \frac{t^{2}}{t^{2}+4} dt \) with respect to \( x \). This involves applying the Fundamental Theorem of Calculus.
2Step 2: Apply the Fundamental Theorem of Calculus
By the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then \( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \). Apply this to both integrals separately.
3Step 3: Differentiate the First Integral
Differentiate \( \int_{-1}^{x} \frac{t^{2}}{t^{2}+4} dt \) with respect to \( x \). Using the Fundamental Theorem of Calculus, the result is simply \( \frac{x^{2}}{x^{2}+4} \).
4Step 4: Differentiate the Second Integral
Differentiate \( \int_{3}^{x} \frac{t^{2}}{t^{2}+4} dt \) with respect to \( x \). Similarly, the result is \( \frac{x^{2}}{x^{2}+4} \) by the Fundamental Theorem of Calculus.
5Step 5: Subtraction of Results
Subtract the result of the second integral from the first: \( \frac{x^{2}}{x^{2}+4} - \frac{x^{2}}{x^{2}+4} \).
6Step 6: Simplify the Expression
Simplify the expression from the subtraction: \( 0 \).

Key Concepts

DifferentiationAntiderivativeIntegral Calculus
Differentiation
Differentiation is a fundamental concept in calculus, focusing on how functions change. It allows us to find the rate at which one quantity changes with respect to another. In simple words, differentiation gives us the derivative of a function.
The derivative of a function at a given point provides a lot of useful information:
  • It tells us the slope of the tangent line to the curve at that point.
  • It indicates how a small change in one variable affects another.
  • It helps in finding maximum and minimum values of functions, which is crucial in optimization problems.
  • It provides insights into the increasing or decreasing nature of a function.
In the given exercise, we use differentiation to find the derivative of a function, which is expressed as an integral. By applying the techniques of differentiation, especially through the Fundamental Theorem of Calculus, we determine how the integral-based function changes with respect to its variable, providing us with the derivative result of zero.
Antiderivative
An antiderivative, also known as an indefinite integral, is essentially the reverse process of differentiation. It refers to a function whose derivative is the given function.
When we talk about finding the antiderivative of a function, we are interested in:
  • Identifying a function that, when differentiated, results in the original function.
  • Adding a constant of integration, because there are infinitely many antiderivatives differing by a constant.
  • Utilizing known antiderivatives to tackle more complex functions.
  • Applying rules like the power rule, sum rule, and integration by parts where necessary.
In the problem we solved, we dealt with finding antiderivatives as part of applying the Fundamental Theorem of Calculus. Though antiderivatives were implicitly involved, the emphasis was more on using our understanding of antiderivatives to compute derivatives of integral expressions.
Integral Calculus
Integral calculus is the branch of calculus that deals with the accumulation of quantities, like areas under curves and the total accumulated change over an interval. It involves two primary types of integrals:
  • Indefinite integrals, which yield antiderivatives.
  • Definite integrals, which provide a numerical value representing the area under a curve across a particular interval.
In the exercise, two definite integrals are presented: - The first integral computes the accumulation from -1 to x. - The second computes from 3 to x.
By evaluating each integral and then finding their difference, we applied integral calculus principles to manipulate and differentiate the expression. This highlights integral calculus's role in solving real-world problems by quantifying changes and accumulations within given boundaries.