Problem 48 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(41-62\). $$ \int_{0}^{\ln 16} e^{x / 4} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 4.

Step 1: Understand the Problem

The integral provided is \( \int_{0}^{\ln 16} e^{x / 4} \, dx \). Our task is to find the antiderivative of the function inside the integral and then evaluate it at the upper and lower limits.

Step 2: Find the Antiderivative

To integrate \( e^{x/4} \), notice that it is a simple exponential function. The antiderivative of \( e^{ax} \) is \( \frac{1}{a}e^{ax} \), where \( a = \frac{1}{4} \) in this case. Thus, the antiderivative is \( 4e^{x/4} \).

Step 3: Apply the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus to evaluate the integral: \( F(b) - F(a) \), where \( F(x) \) is the antiderivative. Here, \( a = 0 \) and \( b = \ln 16 \). Compute \( 4e^{ln 16/4} - 4e^{0/4} \).

Step 4: Simplify the Expression

Calculate \( e^{\ln 16/4} = e^{\ln 2} = 2 \) since \( 16^{1/4} = 2 \). Therefore, \( 4e^{\ln 16/4} = 4 \times 2 = 8 \). Similarly, \( e^{0/4} = e^0 = 1 \).

Step 5: Calculate the Definite Integral

Substitute these values back into the expression to get the solution: \( 8 - 4 \times 1 = 8 - 4 = 4 \). Thus, the value of the integral is 4.

Key Concepts

fundamental theorem of calculus

The Fundamental Theorem of Calculus is a bridge between differentiation and integration. It has two parts, and here we are concerned with using it to evaluate definite integrals. Essentially, this theorem tells us that if you can find an antiderivative of a function, you can calculate the definite integral of that function over an interval

• Find the antiderivative of the function.
• Evaluate it at the upper limit of the interval.
• Subtract its value at the lower limit from its value at the upper limit.

In our example, the integral \(\int_{0}^{\ln 16} e^{x / 4} dx\)is evaluated by first finding its antiderivative. Then, the theorem tells us to take the value of this antiderivative at the upper bound \(\ln 16\)and subtract from it the value at the lower bound \(0\). This gives us the net area under the curve of \(e^{x / 4}\) from \(0\)to \(\ln 16\).

exponential functions

Exponential functions are fundamental in calculus. They have the form \(e^{ax}\), where \(e\) is the base of natural logarithms (approximately 2.718) and \(ax\) is the exponent, which can include any real number.

• They exhibit constant growth or decay, which makes them very predictable and useful in modeling various real-world phenomena.
• In calculus, one special property of the exponential function \(e^{x}\) is how its rate of change is proportional to its current value.

In our integral \(e^{x / 4}\), this type of exponential function has its growth "diluted" by dividing \(x\) by 4. This changes the rate of growth of the function, making it proportional to the expression inside the exponent.

antiderivative

An antiderivative is essentially the reverse of a derivative. If a function \(F(x)\) is an antiderivative of \(f(x)\), then \(F'(x) = f(x)\). This means finding a function whose derivative gives you the function within your integral.

• To find the antiderivative of \(e^{x/4}\), we utilize the fundamental property of the exponential function that the derivative \(d/dx [e^{ax}] = ae^{ax}\).
• This requires dividing \(e^{x/4}\) by \(a = \frac{1}{4}\), resulting in \( 4e^{x/4} \).

This means that \(4e^{x/4}\) is the antiderivative of \(e^{x/4}\).

integration techniques

Integration can involve a variety of techniques depending on the function. Some integrals are straightforward and can be solved by recognizing basic antiderivatives, while others may require more sophisticated methods.

• This particular integral uses a direct approach due to the simplicity of \(e^{x/4}\), a basic exponential function.
• After recognizing the form of the function, one applies the known antiderivative formula for \(e^{ax}\).
• Techniques such as substitution, integration by parts, and partial fractions are not needed here, but are useful for more complex integrals.

Each integral can offer a unique challenge, and understanding which technique to use is crucial for effective integration.