To evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \cos(x) \, dx$ using the Fundamental Theorem of Calculus, let's follow these steps:

Step 1: Find the Antiderivative

The antiderivative of $\cos(x)$ is $\sin(x)$, because: $\frac{d}{dx} \sin(x) = \cos(x).$

Step 2: Apply the Limits of Integration

Using the Fundamental Theorem of Calculus, we evaluate $\sin(x)$ at the upper and lower limits: $\int_{0}^{\frac{\pi}{2}} \cos(x) \, dx = \left[ \sin(x) \right]_{0}^{\frac{\pi}{2}}.$

Step 3: Substitute the Limits

Substitute $x = \frac{\pi}{2}$ and $x = 0$: $= \sin\left(\frac{\pi}{2}\right) - \sin(0).$ Since $\sin\left(\frac{\pi}{2}\right) = 1$ and $\sin(0) = 0$, we get: $= 1 - 0 = 1.$

Final Answer

The value of the integral is: $\int_{0}^{\frac{\pi}{2}} \cos(x) \, dx = 1.0$ to one decimal place.

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Related Questions

1. How does the Fundamental Theorem of Calculus relate antiderivatives to definite integrals?
2. What are the common antiderivatives for trigonometric functions?
3. How would the integral change if the integrand were $\sin(x)$ instead of $\cos(x)$?
4. What is the geometric interpretation of the integral $\int_{0}^{\frac{\pi}{2}} \cos(x) \, dx$?
5. How would you evaluate an integral with different limits, such as $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos(x) \, dx$?

Tip

Remember that the Fundamental Theorem of Calculus allows you to compute definite integrals by finding the antiderivative and applying the limits of integration.