Evaluate the integral. Give result in exact form.
[tex]\int_{\frac{\pi}{2}}^{\frac{5 \pi}{6}}(3 \cos (x)+6) d x[/tex]
Answer (2)
The integral ∫ 2 π 6 5 π ( 3 cos ( x ) + 6 ) d x evaluates to 2 π − 2 3 by finding its antiderivative and applying the Fundamental Theorem of Calculus. The antiderivative is 3 sin ( x ) + 6 x , and we evaluated it at both limits. The final result is obtained by subtracting the value at the lower limit from the value at the upper limit.
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Find the antiderivative of the integrand: 3 sin ( x ) + 6 x .
Evaluate the antiderivative at the upper limit: 3 sin ( 6 5 π ) + 6 ( 6 5 π ) = 2 3 + 5 π .
Evaluate the antiderivative at the lower limit: 3 sin ( 2 π ) + 6 ( 2 π ) = 3 + 3 π .
Subtract the value at the lower limit from the value at the upper limit to get the final answer: 2 π − 2 3 .
Explanation
Problem Setup We are asked to evaluate the definite integral ∫ 2 π 6 5 π ( 3 cos ( x ) + 6 ) d x and express the result in exact form.
Finding the Antiderivative First, we find the antiderivative of the integrand 3 cos ( x ) + 6 . The antiderivative of 3 cos ( x ) is 3 sin ( x ) , and the antiderivative of 6 is 6 x . Therefore, the antiderivative of 3 cos ( x ) + 6 is 3 sin ( x ) + 6 x .
Evaluating at Upper Limit Next, we evaluate the antiderivative at the upper limit of integration, 6 5 π : 3 sin ( 6 5 π ) + 6 ( 6 5 π ) = 3 ( 2 1 ) + 5 π = 2 3 + 5 π .
Evaluating at Lower Limit Then, we evaluate the antiderivative at the lower limit of integration, 2 π : 3 sin ( 2 π ) + 6 ( 2 π ) = 3 ( 1 ) + 3 π = 3 + 3 π .
Subtracting the Values Now, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: ( 2 3 + 5 π ) − ( 3 + 3 π ) = 2 3 + 5 π − 3 − 3 π = 2 π − 2 3 .
Final Answer Therefore, the exact value of the integral is 2 π − 2 3 .
Examples
Imagine you are calculating the total amount of sunlight a solar panel receives over a period of time. If the sunlight intensity is modeled by a cosine function (due to the angle of the sun) plus a constant (representing ambient light), evaluating the definite integral over a specific time interval gives you the total energy received. This is crucial for designing efficient solar energy systems.
