Problem 69 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{0}^{\pi / 4} e^{x} \sin x d x $$

Step-by-Step Solution

Answer

The value of the integral is \(-\frac{1}{2}\).

Step 1: Understand the Integral Problem

We need to evaluate the definite integral \( \int_{0}^{\pi / 4} e^{x} \sin x \; dx \). This integral is a product of an exponential function and a trigonometric function, which suggests using integration by parts.

Step 2: Choose Integration by Parts Components

For integration by parts, choose \( u = e^x \) and \( dv = \sin x \, dx \). Then, differentiate and integrate to find \( du \) and \( v \). We have \( du = e^x \, dx \) and \( v = -\cos x \).

Step 3: Apply Integration by Parts

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du \), substitute the known values:\[\int e^{x} \sin x \, dx = e^x (-\cos x) - \int -\cos x e^x \, dx\] which simplifies to:\[e^x (-\cos x) + \int \cos x e^x \, dx\]

Step 4: Solve the New Integral (Cosine Function)

Now apply integration by parts again to \( \int \cos x e^x \, dx \). Select \( u = e^x \) and \( dv = \cos x \, dx \) which give \( du = e^x \, dx \) and \( v = \sin x \). The integral becomes:\[\int \cos x e^x \, dx = e^x \sin x - \int \sin x e^x \, dx\]

Step 5: Set up and Solve the Equation Involving Both Integrals

Substitute back the new integral from Step 4 into the equation from Step 3:\[\int e^{x} \sin x \, dx = -e^x \cos x + e^x \sin x - \int e^{x} \sin x \, dx\]Add \( \int e^{x} \sin x \, dx \) to both sides to solve for it:\[2\int e^{x} \sin x \, dx = e^x (\sin x - \cos x)\]

Step 6: Find the Indefinite Integral

Solve for the indefinite integral:\[\int e^{x} \sin x \, dx = \frac{1}{2} e^x (\sin x - \cos x) + C\]

Step 7: Evaluate the Definite Integral with Limits

Evaluate from 0 to \( \frac{\pi}{4} \):\[\left[ \frac{1}{2} e^x (\sin x - \cos x) \right]_{0}^{\pi / 4} = \frac{1}{2} \left( e^{\pi / 4} (\sin \frac{\pi}{4} - \cos \frac{\pi}{4}) - e^0 (\sin 0 - \cos 0) \right)\]

Step 8: Simplify the Expression and Calculate

Knowing \( \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), we can simplify:\[= \frac{1}{2} \left( e^{\pi / 4} \cdot (0) - 1(-0 - 1) \right)= \frac{1}{2} (0 - 1)= \frac{1}{2} \cdot (-1)= -\frac{1}{2}\]

Step 9: Finalize and Verify Solution

The evaluated definite integral is \(-\frac{1}{2}\), verifying each step for possible arithmetic and logical errors assures correctness.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to evaluate integrals where direct integration is challenging. When dealing with functions that are products, such as those involving exponential and trigonometric functions, integration by parts can simplify the problem. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] This formula is derived from the product rule of differentiation and requires choosing which part of the integrand to differentiate and which to integrate. A common strategy is to set \( u \) to a function that becomes simpler when differentiated, and \( dv \) to a function that is easy to integrate.

• Choose \( u \) and \( dv \).
• Differentiatiate \( u \) to find \( du \).
• Integrate \( dv \) to find \( v \).

Apply these choices in the integration by parts formula, simplifying your integrals step by step.

Trigonometric Functions

Trigonometric functions are sine, cosine, tangent, and their reciprocals, each playing a crucial role in calculus. In integration by parts, understanding their derivatives and antiderivatives is key. For example, the derivative of \( \sin x \) is \( \cos x \), while the antiderivative of \( \cos x \) is \( \sin x \), and vice versa with an additional negative sign for negative cyclical calculations.

• \( \frac{d}{dx}(\sin x) = \cos x \)
• \( \frac{d}{dx}(\cos x) = -\sin x \)
• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)

The properties and identities of trig functions make them easy to manipulate within integrals, allowing transformations that simplify calculations, especially when paired with other function types.

Exponential Functions

Exponential functions, like \( e^x \), are unique because their rate of growth scales directly with their value. This distinctive feature makes their derivatives and antiderivatives identical in form:

• The derivative: \( \frac{d}{dx}(e^x) = e^x \)
• The integral: \( \int e^x \, dx = e^x + C \)

This property is beneficial when using integration by parts, particularly in problems like \( \int e^x \sin x \, dx \), where the exponential component remains unchanged through differentiation or integration. Combining exponential and trigonometric functions requires careful application of integration techniques, where understanding their respective behaviors—the constant rate of change for \( e^x \) and the cyclical nature of trig functions—facilitates solving complex integrals effectively.