Problem 34 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^3_0 (2\sin x - e^x) \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to approximately -19.10552.

Step 1: State the Integral Problem

We are given the integral \( \int^3_0 (2\sin x - e^x) \,dx \). Our goal is to evaluate this definite integral from 0 to 3.

Step 2: Split the Integral

This integral can be split into two separate integrals: \[ \int_0^3 2\sin x \,dx - \int_0^3 e^x \,dx. \]

Step 3: Integrate Each Function Separately

First, find the antiderivative of each part. 1. For \( 2\sin x \), the antiderivative is \(-2\cos x\).2. For \( e^x \), the antiderivative is \( e^x \).

Step 4: Evaluate the Antiderivatives at the Bounds

Plug in the upper and lower bounds into each antiderivative.1. Evaluate \(-2\cos x\) from 0 to 3: \[ [-2\cos 3] - [-2\cos 0] \approx [-2(-0.98999)] - [-2(1)] = 1.97998 - 2 = -0.02002 \]2. Evaluate \(e^x\) from 0 to 3: \[ [e^3] - [e^0] = e^3 - 1 \approx 20.0855 - 1 = 19.0855 \]

Step 5: Combine the Results

Subtract the two results:\[-0.02002 - 19.0855 = -19.10552.\] This is the evaluated integral of the given function from 0 to 3.

Key Concepts

Antiderivative

An antiderivative is the reverse process of differentiation. When you find the antiderivative of a function, you're looking for a function whose derivative is the original function.
It's useful in calculating integrals which represent the area under a curve between two points.To find the antiderivative, it's important to identify common function patterns. For basic functions like polynomials, trigonometric functions, and exponentials, there are standard antiderivatives:

• The antiderivative of a constant times a function is the constant times the antiderivative of the function.
• For a linear function like \( ax \), it becomes \( ax/2 + C \), where \( C \) is the constant of integration.

In our exercise, we calculated the antiderivative of two functions, \(2\sin x\) and \(e^x\):

• For \(2\sin x\), the antiderivative is \(-2\cos x\) because the derivative of \(-\cos x\) is \(\sin x\).
• For \(e^x\), as its derivative is itself, the antiderivative remains \(e^x\).

These antiderivatives are then evaluated at the given bounds to solve the definite integral.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions like sine, cosine, tangent, etc., which often recur in calculus problems related to periodic phenomena like waves and oscillations.
To integrate trigonometric functions, we often use their known antiderivatives:

• The integral of \(\sin x\) is \(-\cos x + C\).
• The integral of \(\cos x\) is \(\sin x + C\).

In the given exercise, we used the trigonometric integral of \(\sin x\), given that the antiderivative is \(-\cos x\). When multiplying by a constant, such as 2 in \(2\sin x\), you simply multiply the antiderivative by that constant to get \(-2\cos x\). This method provides a straightforward way to integrate trigonometric terms.
Remember, when solving a definite integral, you don't include the constant \(C\) because it cancels out during evaluation of the limits.

Exponential Integrals

Exponential integrals involve integrating functions of the form \(e^x\) or more complex variations like \(e^{kx}\). They are commonly found in growth processes, such as compound interest, population growth, and radioactive decay.
The most common exponential integral is of \(e^x\), whose antiderivative remains \(e^x + C\), due to the unique property of the exponential function that its derivative is the same as the function itself.
In our specific problem, the integral of \(e^x\) does not include any constant multiples, so the antiderivative simply stays as \(e^x\). We evaluate this from the bounds given in the problem. Remember that for any exponential function of the format \(e^{kx}\), the antiderivative would be \(\frac{1}{k}e^{kx} + C\).
This concept is particularly important in solving definite integrals, as the constants cancel when evaluating at the bounds, leading to a neat calculation of the integral's value.