Problem 11 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(7-12,\) use differentiation to verify the antiderivative formula. $$\int \frac{1}{1+u^{2}} d u=\tan ^{-1} u+C$$

Step-by-Step Solution

Answer

By taking the derivative of the result of the integral \(\tan^{-1}(u) + C\) and comparing it to the integrand \(\frac{1}{1+u^2}\), we confirm that the given integral formula holds true.

Step 1: Understand the problem

We are given the integral \(\int \frac{1}{1+u^{2}} d u=\tan ^{-1} u+C\) and are asked to verify this equality using differentiation. If we take derivative of the right side of the equation, we should end up with the integrand on the left side, as per the Fundamental Theorem of Calculus.

Step 2: Differentiate the antiderivative

Differentiate \(\tan^{-1}(u) + C\) using the chain rule. The derivative of \(\tan^{-1}(u)\) is \(\frac{1} {1+u^2} \), and the derivative of a constant \(C\) is zero. Thus, the derivative of \(\tan^{-1}(u) + C\) is \(\frac{1} {1+u^2} \).

Step 3: Verify the equality

Comparing this result with the integrand of the given integral, we notice that they are identical. This confirms that \(\tan^{-1}(u) + C\) is indeed an antiderivative of \(\frac{1}{1+u^2}\).

Key Concepts

Antiderivative

An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function. In simple terms, if you have a function \(f(x)\), finding its antiderivative means identifying a function \(F(x)\) such that \(F'(x) = f(x)\).An important property of antiderivatives is that they are not unique. They can differ by a constant.When calculating antiderivatives, you often see the notation \(\int f(x) \, dx = F(x) + C\), where \(C\) is an arbitrary constant to account for any initial differences.

• For example, the antiderivative of \(\frac{1}{1+u^2}\) is \(\tan^{-1}(u) + C\).
• The role of \(C\) is crucial because it makes the antiderivative accurate for any potential shifts in the original function.

It's always important to remember to include this constant \(C\) when presenting the result of an indefinite integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus creates a link between differentiation and integration, two primary concepts in calculus.It's split into two parts:

• First Part: It tells us that the integral of a function over an interval can be found using its antiderivative.If \(F\) is an antiderivative of \(f\) on an interval \([a, b]\), then:\[\int_a^b f(x) \, dx = F(b) - F(a)\]
• Second Part: This part states that if you take the derivative of an integral, you get the original function back.Mathematically speaking, if \(F(x) = \int_a^x f(t) \, dt\), then \(F'(x) = f(x)\).

In the context of the exercise, the equation \(\int \frac{1}{1+u^{2}} du = \tan^{-1}(u) + C\) is verified using differentiation.By showing that the derivative of \(\tan^{-1}(u)\) matches the integrand \(\frac{1}{1+u^2}\), we've utilized the second part of this theorem effectively.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions.A composite function is one where you have one function inside of another function, like \(f(g(x))\).The chain rule states that to differentiate a composite function, you need to:

• Find the derivative of the outer function \(f\) evaluated at the inner function \(g(x)\).
• Multiply it by the derivative of the inner function \(g(x)\).

We can write it mathematically as:\[( f(g(x)) )' = f'(g(x)) \cdot g'(x)\]In the provided exercise, the chain rule is applied indirectly.Differentiating \(\tan^{-1}(u)\) highlights the essence of the chain rule, although it might not seem obvious at first.In this case, since \(u\) is the variable and not a function of another variable, the inner function's derivative \(g'(u)\) is simply \(1\).Thus, the derivative of \(\tan^{-1}(u)\) with respect to \(u\) simplifies to \(\frac{1}{1+u^{2}}\), confirming that it is an antiderivative of the provided integrand.