Problem 108 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{0}^{1} \sin (\pi(x+1)) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{-2}{\pi} \).

Step 1: Identify the Substitution

Consider the integral \( \int_{0}^{1} \sin (\pi(x+1)) \ dx \). To simplify this, we'll first make a substitution. Let \( u = \pi(x+1) \). Hence, the derivative \( du = \pi \, dx \). Therefore, \( dx = \frac{du}{\pi} \).

Step 2: Change the Limits

With our substitution \( u = \pi(x+1) \), we need to change the limits of integration. When \( x = 0 \), \( u = \pi(0+1) = \pi \). When \( x = 1 \), \( u = \pi(1+1) = 2\pi \). So the definite integral in terms of \( u \) will now be from \( \pi \) to \( 2\pi \).

Step 3: Substitute and Simplify the Integral

The integral becomes \( \int_{\pi}^{2\pi} \sin(u) \cdot \frac{1}{\pi} \, du \), which simplifies to \( \frac{1}{\pi} \int_{\pi}^{2\pi} \sin(u) \, du \).

Step 4: Evaluate the Integral

The antiderivative of \( \sin(u) \) is \( -\cos(u) \). Therefore, we need to evaluate \( \frac{1}{\pi} \left[ -\cos(u) \right]_{\pi}^{2\pi} \).

Step 5: Calculate the Definite Integral

Substitute the limits into the antiderivative: \( \frac{-1}{\pi} ([\cos(2\pi) - \cos(\pi)]) \). Since \( \cos(2\pi) = 1 \) and \( \cos(\pi) = -1 \), this becomes \( \frac{-1}{\pi} (1 - (-1)) = \frac{-1}{\pi} \times 2 = \frac{-2}{\pi} \).

Step 6: Final Result

The evaluated definite integral, \( \int_{0}^{1} \sin (\pi(x+1)) \, dx \), is \( \frac{-2}{\pi} \).

Key Concepts

Substitution Method

When tackling integrals, especially those involving composite functions or products, the substitution method is a powerful tool. It allows us to simplify an integral by making a substitution that transforms it into a more recognizable form. In our example, we recognized that the function inside the sine, \( \sin(\pi(x+1)) \), was complex. To simplify, we opted for the substitution \( u = \pi(x+1) \).

• This turned the integral into something easier to evaluate by standard trigonometric techniques.
• The derivative \( du = \pi \, dx \) was used to express \( dx \) in terms of \( du \), giving \( dx = \frac{du}{\pi} \).
• After substitution, we switched the problem from the variable \( x \) to the new variable \( u \), making computation much simpler.

Substitution is especially useful in definite integrals because it not only simplifies the function but also adjusts the integration limits to match the new variable, making the problem more manageable for integration.

Trigonometric Integrals

Trigonometric integrals often involve functions like sine, cosine, tangent, and their inverses. These require specific strategies for integration, many of which are rooted in recognizing patterns or applying identities. In our integral \( \int \sin(u) \, du \), we made use of trigonometric properties.

• The antiderivative of \( \sin(u) \) is \( -\cos(u) \), a fundamental result that simplifies many problems.
• Understanding these basic trigonometric antiderivatives is crucial, as it forms the foundation for tackling more complex integrals.
• In problems where the integrals of sine and cosine disturb the regular input, substitution often leads directly to these simpler forms.

Thus, knowing the basic forms and properties of trigonometric functions and their integrals enables us to evaluate them confidently and correctly.

Limits of Integration

Definite integrals have boundaries, known as limits of integration, which mark the range over which the area under the curve is calculated. When performing substitution, it is vital to adjust these limits to the new variable. For our problem, the original limits were from \( x = 0 \) to \( x = 1 \).

• We used the substitution \( u = \pi(x+1) \) to find new limits.
• When \( x = 0 \), \( u = \pi \), and when \( x = 1 \), \( u = 2\pi \).
• By adjusting these limits, the new integral correctly accounts for the area between the transformed bounds \( \pi \) and \( 2\pi \).

This modification in limits ensures the integration is carried out within the correct scope, thereby providing an accurate result that reflects the initial problem conditions.