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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.

Would you like further details or have any questions?

### 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.

Evaluate the definite integral ∫ from 0 to π/2 of cos(x) dx using the Fundamental Theorem of Calculus. Leave your answer to 1 decimal place.

Learn how to evaluate the definite integral of cos(x) from 0 to π/2 using the Fundamental Theorem of Calculus. This problem demonstrates the integration of trigonometric functions and is suitable for students in Grades 11-12.

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