Problem 30 - Applied Calculus

Question

Use an integral to find the specified area. Between \(y=\sin x+2\) and \(y=0.5\) for \(6 \leq x \leq 10\).

Step-by-Step Solution

Answer

The area is approximately 6.121 square units.

Step 1: Understand the Problem

We need to find the area between a curve and a horizontal line over a specified interval. The functions given are \(y = \sin x + 2\) and \(y = 0.5\). The interval is for \(x\) values from 6 to 10.

Step 2: Set up the Integral

The area between the graphs of two functions \(f(x)\) and \(g(x)\) over an interval \([a, b]\) is found by computing the definite integral of the absolute difference of these functions: \( \int_a^b |f(x) - g(x)| \, dx \). In this case, \(f(x) = \sin x + 2\) and \(g(x) = 0.5\), with limits \(a = 6\) and \(b = 10\). This gives us the integral \( \int_6^{10} ((\sin x + 2) - 0.5) \, dx \).

Step 3: Simplify the Integral

Simplify the integrand: \((\sin x + 2) - 0.5 = \sin x + 1.5\). Thus, the integral becomes \(\int_6^{10} (\sin x + 1.5) \, dx\).

Step 4: Compute the Integral

Integrate \( \sin x + 1.5\) with respect to \(x\):\[ \int (\sin x + 1.5) \, dx = -\cos x + 1.5x + C \]Evaluate this antiderivative from 6 to 10:\[ \left[ -\cos x + 1.5x \right]_6^{10} = \left(-\cos(10) + 1.5 \times 10\right) - \left(-\cos(6) + 1.5 \times 6\right) \]

Step 5: Evaluate the Expression

Substitute the values into the expression:- First for \(x = 10\): -\(-\cos(10) + 15 = -\cos(10) + 15\)- Then for \(x = 6\): -\(-\cos(6) + 9 = -\cos(6) + 9\)Calculate the differences: \[-\cos(10) + 15 - (-\cos(6) + 9)\] = \[-\cos(10) + \cos(6) + 6\].

Step 6: Final Calculation

Use a calculator to approximate \(-\cos(10) \approx -0.8391\) and \(-\cos(6) \approx -0.9601\). Thus:\[ -0.8391 + 0.9601 + 6 = 0.121 + 6 = 6.121\].

Key Concepts

Area between curves

Finding the area between two curves is a common application of definite integrals in calculus. When asked to find the area between two curves, essentially, you need to determine the region bounded by these curves. In the given exercise, the two functions are \(y = \sin x + 2\) and \(y = 0.5\). These create a specific shape between the given interval \([6, 10]\).

To find this area, you’ll need to integrate the difference of these functions over the specified interval. The integral used is \[\int_a^b |f(x) - g(x)| \, dx\] where \(f(x)\) is the upper function, and \(g(x)\) is the lower function within that region.

• Identify the upper and lower functions in the region. Here, \(\sin x + 2\) is above \(0.5\) for \(x\) in \([6, 10]\).
• Subtract the lower function from the upper one, ensuring the expression inside the integral is positive across the integration interval.

This method allows you to compute the exact total area between the curves.

Trigonometric integration

Trigonometric integration involves integrating functions that contain trigonometric functions like sine or cosine, which is a crucial aspect of calculus. In this scenario, the function \(f(x) = \sin x + 2\) requires us to apply integration techniques to trigonometric functions as part of the process to find the area.

When integrating a sine function like \(\sin x\), the antiderivative is \(-\cos x\). This arises from the derivative rules for trigonometric functions. The integration of constants, like \(2\), results in the multiplication with the variable, so the antiderivative of \(2\) is \(2x\).

• For \(\sin x\), the integration is \(-\cos x\).
• For a constant \(2\), integrate to get \(2x\).
• Combining, the integral of \(\sin x + 2\) becomes \(-\cos x + 2x\).

These integrals accumulate to give us the expression we need to evaluate from 6 to 10.

Evaluation of definite integral

The evaluation of definite integrals involves much more than just setting up an expression. It’s a process where you compute the net area under a curve within a specified interval, using antiderivatives.

Once the integral expression is found, such as \[\int_6^{10} (\sin x + 1.5) \, dx\], applying the antiderivative calculations followed by evaluating this at the bounds is essential. Here's the step-by-step breakdown:

• **Find the antiderivative**: For the expression, this results in \(-\cos x + 1.5x\).
• **Evaluate at upper bound \(x=10\)**: Substitute \(x = 10\) into the antiderivative to calculate this part.
• **Evaluate at lower bound \(x=6\)**: Do the same for \(x = 6\).

Subtract the lower bound evaluation from the upper bound evaluation to get the net area. This results in \[-\cos(10) + 15 - (-\cos(6) + 9)\] simplifying to \[-\cos(10) + \cos(6) + 6\].

Finally, calculate these using a calculator to obtain an approximate numeric result, as shown in the solution: \(6.121\).