Problem 418 - Calculus Volume 1

Question

In the following exercises, use a calculator to graph the antiderivative \(\int f\) with \(C=0\) over the given interval \([a, b]\) . Approximate a value of \(C,\) if possible, such that adding \(C\) to the antiderivative gives the same value as the definite integral \(F(x)=\int_{a}^{x} f(t) d t\). $$[\mathbf{T}] \int \frac{1}{(2 x+2) \sqrt{x}} d x \text { over }[0,6]$$

Step-by-Step Solution

Answer

With \( C = 0 \), the antiderivative yields the same value as the definite integral over \([0,6]\).

Step 1: Break Down the Integral

The problem is asking you to find the antiderivative of the function \( f(x) = \frac{1}{(2x+2)\sqrt{x}} \) over the interval \([0, 6]\). Let's first consider simplifying the expression before attempting to integrate. We will focus on finding the antiderivative analytically and evaluate it at the bounds \( a \) and \( b \).

Step 2: Simplify the Function

Rewrite the function as \( f(x) = \frac{1}{(2x+2)\sqrt{x}} = \frac{1}{2(x+1)\sqrt{x}} \). Now, let's consider a substitution to simplify this even further. Using \( u = \sqrt{x} \), then \( du = \frac{1}{2\sqrt{x}}dx \rightarrow dx = 2u du \). Thus, \( x = u^2 \), and the function becomes \( \frac{2u}{2(u^2 + 1)u}du = \frac{1}{u^2 + 1}du \).

Step 3: Integrate Using Substitution

Now integrate \( \frac{1}{u^2 + 1} \), which is \( \int \frac{1}{u^2 + 1} \, du = \arctan(u) + C \). Since \( u = \sqrt{x} \), the antiderivative becomes \( \arctan(\sqrt{x}) + C \).

Step 4: Evaluate Definite Integral

Using the antiderivative \( F(x) = \arctan(\sqrt{x}) \), calculate \( F(6) - F(0) \) to find the integral over [0, 6]. This gives \( \arctan(\sqrt{6}) - \arctan(\sqrt{0}) = \arctan(\sqrt{6}) \).

Step 5: Determine the Appropriate Constant C

Given that \( C = 0 \) in the indefinite integral for matching the value with the definite integral, any constant \( C \) needs circumstances where \( F(x) + C = F(6) \). Since the derivative did not have an arbitrary constant initially, we conclude that adding \( C = 0 \) fits the requirement as any constant adjustment would not fulfill evaluating the definite integral requirements seamlessly.

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus, as it helps compute the accumulation of quantities, like areas under curves. The definite integral is represented by the expression \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration, serving as the lower and upper bounds of the interval. In essence, this is calculating the total net area under the graph of \( f(x) \) from \( x = a \) to \( x = b \).

• It requires the function \( f(x) \) to be properly behaved within the interval \([a, b]\).
• The result is a numerical value, representing the accumulated effect of the function across the interval.
• This approach is crucial when evaluating physical quantities like displacement when given a velocity-time graph.

Understanding definite integrals also gives rise to the concept of an antiderivative or a function whose derivative is the integrand. Thus, accurately determining a definite integral often involves finding or knowing the antiderivative of a function.

Substitution Method

The substitution method in integration, much like its counterpart in the derivative, is a technique used to simplify more complex integrals. This method involves changing the variables in the integral to make it easier to solve. In the substitution method, we set a part of the integral equal to a new variable, usually called \(u \), thereby transforming the integral into a simpler form.

• Choose a substitution that simplifies the integral, by substituting \( u = g(x) \) such that \( du = g'(x)dx \).
• Rewrite the differential \( dx \) in terms of the new variable \( du \).
• Transform the integral into \( \int f(u) \, du \) and solve.

For instance, consider \( f(x) = \frac{1}{(2x+2)\sqrt{x}} \):

• By substituting \( u = \sqrt{x} \), we ease computation as \( dx = 2u \, du \) transforms the integral to a recognizable format.
• This results in \( \int \frac{1}{u^2 + 1} \, du \), which is much more straightforward to integrate.

Using substitutions not only eases computation but also aids in understanding underlying patterns and transformations in functions.

Interval Evaluation

Evaluating an integral over a specific interval is where the transition from indefinite to definite integrals occurs. It involves calculating the integral's value between two bounds, thereby determining the total change or effect of a function on that interval.

• Compute the antiderivative \( F(x) \) of the integrand \( f(x) \).
• Substitute the upper bound into the antiderivative to get \( F(b) \) and the lower bound to get \( F(a) \).
• The definite integral is then \( F(b) - F(a) \).

In our specific example, the antiderivative became \( \arctan(\sqrt{x}) \).

• Calculating at the bounds yields \( \arctan(\sqrt{6}) - \arctan(\sqrt{0}) = \arctan(\sqrt{6}) \).
• Such evaluations are helpful for deriving numerical results that relate to real-world problems, like the area under a function representing velocity or other rates of change.

Interval evaluation is thus both a conceptual and practical tool, providing critical insights into the behavior of functions between specified points.

Constant of Integration

In any indefinite integral, the constant of integration \( C \) plays an essential role. Initially, when finding antiderivatives, you include this constant to account for any shift in the vertical position of the graph of the function, as the derivative of a constant is zero.

• The constant \( C \) ensures all potential antiderivatives are represented, as each could differ by a constant factor.
• For definite integrals, \( C \) is not necessary when evaluating limits, as it cancels itself out when calculating \( F(b) - F(a) \).
• In practice, finding the right constant may require additional conditions from context-specific scenarios to determine precisely or ensure continuity of a solution.

In the exercise, \( C = 0 \) was chosen so that the value of the indefinite integral aligned correctly when compared directly to the definite integral result. While for indefinite integrals the constant is crucial, for definite integrals, it often plays a minimal role in altering numerical outcomes from specific interval evaluations.