Determine the area of the shaded region in the following figures.

검증된 단계별 안내

Identify the curves that bound the shaded region. Here, the region is bounded by the curves \(y = \sin x\) and \(y = \cos x\) between \(x = 0\) and \(x = \frac{\pi}{2}\).

Determine which function is on top and which is on the bottom in the interval \([0, \frac{\pi}{2}]\). Since \(\cos x\) starts at 1 and decreases, and \(\sin x\) starts at 0 and increases, \(\cos x\) is above \(\sin x\) in this interval.

Set up the integral for the area of the shaded region as the integral of the difference between the top function and the bottom function over the interval: \(\int_0^{\frac{\pi}{2}} (\cos x - \sin x) \, dx\).

Integrate the function \(\cos x - \sin x\) with respect to \(x\). Recall that the integral of \(\cos x\) is \(\sin x\) and the integral of \(\sin x\) is \(-\cos x\).

Evaluate the definite integral by substituting the limits \(0\) and \(\frac{\pi}{2}\) into the antiderivative and subtracting to find the exact area of the shaded region.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Definite Integrals and Area Between Curves

The area between two curves over an interval can be found using definite integrals by integrating the difference of the functions. Specifically, the area is the integral of the upper function minus the lower function from the left to the right boundary of the region.

Trigonometric Functions and Their Properties

Understanding the behavior of sine and cosine functions, including their values and intersections within the interval [0, π/2], is essential. Knowing where sin x and cos x intersect helps determine the limits and which function is on top in the region.

Setting Up the Integral Limits

Identifying the correct interval for integration is crucial. Here, the shaded region lies between x = 0 and x = π/2, where the two curves intersect and bound the area. Properly setting these limits ensures accurate calculation of the shaded area.