91Ó°ÊÓ

A definite integral is a key concept in calculus that helps calculate the area between two curves, among other things. Essentially, it represents the signed area under a curve from one point to another. For this exercise, we evaluate the definite integral of the difference between two functions over a specified interval.
To solve the given problem, we first identify the curves and the interval. We then set up the integral properly. In our example, we have:

• Linear function: \( y = x + 3 \)
• Trigonometric function: \( y = \sin x + \cos x \)
• Interval: \( x = 0 \) to \( x = \frac{\pi}{2} \)

To find the area between these curves over the specified interval, the definite integral is set up as: \[ A = \int_{0}^{\pi/2} \big[ (x + 3) - (\sin x + \cos x) \big] \, dx \] This calculation will give us the area of the region bounded by these graphs.

Trigonometric functions arise frequently in calculus problems. In our exercise, we have the function \( y = \sin x + \cos x \). Here’s a brief refresher on these functions:

• Sine (\( \sin x \)): This function returns the ratio of the length of the opposite side to the length of the hypotenuse in a right-angled triangle.
• Cosine (\( \cos x \)): This function returns the ratio of the length of the adjacent side to the length of the hypotenuse in a right-angled triangle.

When combined as \( y = \sin x + \cos x \), integrating this function involves understanding their periodicity and the rules for integrating trigonometric functions. For example:

• \( \int \sin x \,dx = -\cos x \bigg|_0^{\pi/2} \) returns 1
• \( \int \cos x \,dx = \sin x \bigg|_0^{\pi/2} \) also returns 1.

Knowing how to manipulate and work with these functions is crucial for solving calculus problems.

Various integration techniques help us evaluate integrals effectively. In our current exercise, we need to integrate both polynomial and trigonometric functions over a defined interval. Here’s a step-by-step on the techniques used in our exercise:

• Integration of a Polynomial: For \( f(x) = x \), the integral is \( \int x \,dx = \frac{x^2}{2} \). For a constant 3, it’s \( \int 3 \,dx = 3x \).
• Integration of Trigonometric Functions: For \( \sin x \), it’s \( -\cos x \), and for \( \cos x \), it’s \( \sin x \).

We evaluate these integrals separately over the interval \( [0, \frac{\pi}{2}] \). Combining these results gives the final area. Here are the integrals evaluated:

• \( \int_{0}^{\pi/2} x \, dx = \frac{(\frac{\pi}{2})^2}{2} = \frac{\pi^2}{8} \)
• \( \int_{0}^{\pi/2} 3 \, dx = 3 \frac{\pi}{2} = \frac{3\pi}{2} \)
• \( \int_{0}^{\pi/2} \sin x \, dx = [ - \cos x ]_0^{\pi/2} = 1 \)
• \( \int_{0}^{\pi/2} \cos x \, dx = [ \sin x ]_0^{\pi/2} = 1 \)

By combining these results, we simplify the final area calculation to: \[ A = \frac{\pi^2}{8} + \frac{3\pi}{2} - 2 \] Using the right techniques makes integrating even complex functions manageable!