Problem 18 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{5}\left(2 e^{x}+4 \cos x\right) d x\)

Step-by-Step Solution

Answer

The integral evaluates to \(2(e^5 - 1) + 4\sin 5\).

Step 1: Break Down the Integral

The given integral is \( \int_{0}^{5}(2e^{x} + 4\cos x)\, dx \). This can be expressed as two separate integrals: \( \int_{0}^{5} 2e^{x} \, dx + \int_{0}^{5} 4\cos x \, dx \). This step aims to simplify the problem by handling each part separately.

Step 2: Solve the First Integral

For \( \int_{0}^{5} 2e^{x} \, dx \), we can factor out the constant to get \( 2 \int_{0}^{5} e^{x} \, dx \). The integral of \( e^{x} \) is itself \( e^{x} \), so we have \( 2 \left[ e^{x} \right]_{0}^{5} \). Evaluate to get \( 2 (e^5 - e^0) = 2 (e^5 - 1) \).

Step 3: Solve the Second Integral

For \( \int_{0}^{5} 4 \cos x \, dx \), factor out the constant to get \( 4 \int_{0}^{5} \cos x \, dx \). The integral of \( \cos x \) is \( \sin x \), so we have \( 4 \left[ \sin x \right]_{0}^{5} \). Evaluate to get \( 4 (\sin 5 - \sin 0) = 4 \sin 5 \).

Step 4: Add the Results

Combine the results of the two integrals to find the final value. The result is \( 2(e^5 - 1) + 4\sin 5 \).

Step 5: Final Answer

The evaluated integral of \( \int_{0}^{5}(2e^{x} + 4\cos x)\, dx \) is \( 2(e^5 - 1) + 4\sin 5 \).

Key Concepts

Definite Integrals

A definite integral is an integral with specified upper and lower limits. It's used to calculate the net area under a curve within a given interval. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.

• The lower limit is the point at which the calculation starts.
• The upper limit is where the calculation stops.

To solve a definite integral, we find the antiderivative of the function, which is the reverse operation of a derivative. We then evaluate this antiderivative at both the upper and lower limits, and subtract the two results. For instance, if we have an integral with limits from 0 to 5, we find the antiderivative, evaluate it at 5 and at 0, and subtract: \( F(5) - F(0) \). This calculation corresponds to the exact area under the curve, giving us the total effect within that interval.

Integration Techniques

Integration techniques are methods used to find the antiderivative of functions. These techniques can simplify the process of solving integrals, especially when the function is complex. Two common techniques are:

• Breaking Down the Integral: This involves splitting a complex integral into simpler parts that can be handled individually. This was used in the exercise where the integral was split into two: \( \int_{0}^{5} 2e^{x} \, dx + \int_{0}^{5} 4\cos x \, dx \).
• Constant Multiplication: Factoring constants out of the integral makes it easier. For example, in the exercise, constants were factored out as \( 2 \int e^{x} \, dx \) and \( 4 \int \cos x \, dx \).

These methods simplify the integration process and allow the use of known antiderivatives, making it straightforward to evaluate complex expressions.

Exponential Functions

Exponential functions often appear in calculus and are characterized by a constant base raised to a variable exponent. A common exponential function to integrate is \( e^{x} \), where \( e \) is the mathematical constant approximately equal to 2.718.

• The antiderivative of \( e^{x} \) is \( e^{x} \), which simplifies the integration process significantly.
• When integrating a function like \( 2e^{x} \), we can factor the constant 2 out, making it \( 2 \int e^{x} \, dx \) and then integrate \( e^{x} \, dx \).
• The calculation becomes even simpler because the antiderivative has a straightforward format, directly leading to the evaluation at the limits of integration.

Exponential functions' properties make them fundamentally predictable and manageable within integral calculus.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are prevalent in calculus. Integrating these functions requires understanding their properties and antiderivatives.

• For \( \cos x \), the antiderivative is \( \sin x \).
• In our exercise, after factoring out the constant and solving \( \int \cos x \, dx \), we have the result \( \sin x \), which is then evaluated over the given limits.
• For trigonometric functions, it's beneficial to be familiar with the fundamental identities and their derivatives, as this knowledge simplifies integration.

By using these identities, we can integrate trigonometric functions more easily, anticipating their behavior and evaluating their impacts within definite integrals.