Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{0}^{1}\left(e^{2 x}+x^{2}+1\right) d x$$

Step-by-Step Solution

Answer

The value of the definite integral \(\int_{0}^{1}\left(e^{2 x}+x^{2}+1\right) d x\) is \(\frac{1}{2} e^2 + \frac{1}{3}\).

Step 1: Identify the antiderivative of each part of the function

We have three parts in our integral function: \(e^{2x}\), \(x^2\), and 1. First, let's find the antiderivatives for each of these parts. 1. Antiderivative of \(e^{2x}\): Using the rule for exponential functions, the antiderivative of \(e^{2x}\) is \(\frac{1}{2} e^{2x}\). 2. Antiderivative of \(x^2\): Using the power rule, the antiderivative of \(x^2\) is \(\frac{1}{3} x^3\). 3. Antiderivative of 1: The antiderivative of a constant is just the constant multiplied by x. So, the antiderivative of 1 is x. Now that we have the antiderivatives of each part, we can write the antiderivative of the entire function.

Step 2: Construct the antiderivative of the entire function

Combine the antiderivatives of each part to form the antiderivative of the whole function F(x): F(x) = \(\frac{1}{2} e^{2x} + \frac{1}{3} x^3 + x\)

Step 3: Evaluate the antiderivative at the limits

Now, we need to evaluate F(x) at the given limits (0 and 1) and subtract the results. This will give us the value of the definite integral. F(1) = \(\frac{1}{2} e^{2(1)} + \frac{1}{3} (1)^3 + (1) = \frac{1}{2} e^2 + \frac{1}{3} + 1\) F(0) = \(\frac{1}{2} e^{2(0)} + \frac{1}{3} (0)^3 + (0) = \frac{1}{2} e^0 = \frac{1}{2}\) Finally, subtract F(0) from F(1) to find the value of the integral: \(\int_{0}^{1}\left(e^{2 x}+x^{2}+1\right) d x = F(1) - F(0) = \left(\frac{1}{2} e^2 + \frac{1}{3} + 1\right) - \frac{1}{2} = \frac{1}{2} e^2 + \frac{1}{3}\) The value of the definite integral is \(\frac{1}{2} e^2 + \frac{1}{3}\).

Key Concepts

Antiderivative

An antiderivative, sometimes called an indefinite integral, is a function whose derivative gives back the original function. Finding an antiderivative is like doing the reverse of differentiation. In the context of definite integrals, as seen in our exercise, we find the antiderivative to then evaluate it over a specific interval.

For instance:

• For the function \( e^{2x} \), the antiderivative is \( \frac{1}{2} e^{2x} \). Here, a multiplying factor comes into play because of the chain rule in reverse.
• For \( x^2 \), the antiderivative is achieved using the power rule which results in \( \frac{1}{3} x^3 \).
• Lastly, for the constant function, 1, the antiderivative is simply \( x \) since the derivative of \( x \) is 1.

Knowing how to find the antiderivative accurately is crucial for evaluating integrals and solving various calculus problems.

Exponential Function

Exponential functions are mathematical functions of the form \( e^{kx} \), where \( e \) is Euler's number (approximately 2.71828) and \( k \) is a constant.

These functions have unique properties. Their rate of change is directly proportional to their value, making them critical in fields like growth phenomena and decay processes.

When integrating exponential functions like \( e^{2x} \), it's essential to adjust for the constant \( k \) in the exponent:

• The antiderivative of \( e^{2x} \) is \( \frac{1}{2} e^{2x} \). The \( \frac{1}{2} \) comes from dividing by the constant \( 2 \) (from the exponent), which is derived from the chain rule reversed.

Exponential functions simplify integration as their derivatives and antiderivatives maintain a similar form, aside from constant factors.

Power Rule

The power rule for integration is a straightforward technique, particularly useful when dealing with polynomial functions.

It states that for any power of \( x \) (except \( x^{-1} \)), you increase the exponent by 1 and then divide by the new exponent.

• For example, given \( x^n \), the antiderivative is \( \frac{1}{n+1} x^{n+1} \).
• In our exercise, we used the power rule to find the antiderivative of \( x^2 \) to be \( \frac{1}{3} x^3 \), since \( n = 2 \).

The power rule is fundamental in calculus as it lets us easily find antiderivatives for a broad class of functions.

Integration Limits

Integration limits are the bounds within which we evaluate the integral. When dealing with definite integrals, as in our task, these limits give the interval where we need to calculate the antiderivative.

• In our example, the integration limits are 0 and 1, meaning we evaluate the antiderivative at both points.
• The process follows these steps: Calculate \( F(x) \) at the upper limit, subtract the value of \( F(x) \) at the lower limit.

This results in the definite integral value. Integration limits are crucial because they provide the specific segment over which we apply the integral, converting an antiderivative into a definite, numeric result.