Problem 60 - Calculus Early Transcendentals

Question

Find the value of \(c\) such that the region bounded by \(y=c \sin x\) and the \(x\) -axis on the interval \([0, \pi]\) has area 1.

Step-by-Step Solution

Answer

Answer: The value of \(c\) that makes the area under the curve equal to 1 is \(c = \frac{1}{2}\).

Step 1: Integrate the function over the interval

We need to find the integral of the function \(y = c \sin x\) with respect to \(x\) on the interval \([0, \pi]\). This will give us the total area under the curve between these points: $$ \int_{0}^{\pi} c \sin x \, dx $$

Step 2: Apply the integration properties

The integral of a constant \(c\) multiplied by a function is equal to the constant multiplied by the integral of the function. Using this property, we can simplify the integral as follows: $$ c \int_{0}^{\pi} \sin x \, dx $$

Step 3: Find the antiderivative of the function

The antiderivative of \(\sin x\) is \(-\cos x\). Therefore, we have: $$ c \left[-\cos x\right]_{0}^{\pi} $$

Step 4: Evaluate the integral

Now we need to evaluate the antiderivative at the limits of the interval and subtract: $$ c(-\cos \pi - (-\cos 0)) = c(1 - (-1)) = 2c $$

Step 5: Solve for \(c\)

We want the area under the curve to be equal to 1. So, we can set up the equation: $$ 2c = 1 $$ Now we just need to solve for \(c\): $$ c = \frac{1}{2} $$

Step 6: Final answer

The value of \(c\) for which the area under the curve \(y = c \sin x\) on the interval \([0, \pi]\) is equal to 1 is \(c = \frac{1}{2}\).

Key Concepts

Definite Integral

In integral calculus, one powerful tool we frequently use is the definite integral. A definite integral calculates the net area between a function and the x-axis over a specific interval. For example, by integrating the function \(y = c \sin x\) from \(0\) to \(\pi\), we determine the area occupied by this curve within this range.

To perform definite integration, we follow these general steps:

• Identify the function to integrate and its limits of integration.
• Find the antiderivative (or indefinite integral) of the function.
• Evaluate the antiderivative at the upper and lower limits of integration.
• Subtract these evaluations to obtain the integral's value.

In our example, calculating the definite integral involves evaluating \(y = c \sin x\) over the interval \([0, \pi]\). Applying these steps ensures we accurately compute the area under the curve.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and tangent, play a crucial role in calculus. They often appear in problems involving periodic behavior and oscillations, such as waves. The sine function, \(\sin x\), maps angles to the vertical sides of a right triangle or provides the y-coordinates of points on a unit circle.

When working with such functions in calculus, understanding their basic properties is essential. For instance:

• \(\sin(0) = 0\) and \(\sin(\pi) = 0\), indicating zero values at these notable points.
• The antiderivative of \(\sin x\) is \(-\cos x\), enabling us to compute integrals involving sine effectively.
• Sine has a regular pattern, repeating every \(2\pi\) radians, useful in determining periodical interval results.

In our exercise, \(\sin x\) is utilized in deriving the area under the curve \(y = c \sin x\), illustrating its application in integration.

Area Under Curve

Finding the area under a curve is a common problem in calculus, often solved using the definite integral. This method provides a systematic way to measure regions bounded by a curve and the x-axis over a specified interval.

To determine this area for a function like \(y = c \sin x\):

• We first express the area as a definite integral, e.g., \(\int_{0}^{\pi} c \sin x \, dx\).
• Next, we calculate the antiderivative, then evaluate it at the boundaries of the given interval.
• Finally, these values are subtracted to find the total area.

In our example, seeking an area of 1 led to solving \(2c = 1\) for \(c\). This showed how integrals can scale areas by altering function coefficients like \(c\), crucial in numerous applications such as physics or engineering.