Evaluate the integral. Give result in exact form.
[tex]\int_{\frac{\pi}{2}}^{\frac{5 \pi}{6}}(11 \cos (x)+3) d x=[/tex]
Answer (2)
To evaluate the integral ∫ 2 π 6 5 π ( 11 cos ( x ) + 3 ) d x , find the antiderivative 11 sin ( x ) + 3 x and evaluate it at the limits. After evaluation and subtraction, the result is π − 2 11 .
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Find the antiderivative of the integrand: 11 sin ( x ) + 3 x .
Evaluate the antiderivative at the upper limit: 2 11 + 2 5 π .
Evaluate the antiderivative at the lower limit: 11 + 2 3 π .
Subtract the value at the lower limit from the value at the upper limit and simplify: π − 2 11 .
Explanation
Problem Setup We are asked to evaluate the definite integral: ∫ 2 π 6 5 π ( 11 cos ( x ) + 3 ) d x This means we need to find the antiderivative of the integrand, and then evaluate it at the upper and lower limits of integration, and finally subtract the value at the lower limit from the value at the upper limit.
Finding the Antiderivative First, let's find the antiderivative of 11 cos ( x ) + 3 . The antiderivative of cos ( x ) is sin ( x ) and the antiderivative of 3 is 3 x . Therefore, the antiderivative of 11 cos ( x ) + 3 is 11 sin ( x ) + 3 x .
Evaluating at Upper Limit Next, we evaluate the antiderivative at the upper limit of integration, 6 5 π :
11 sin ( 6 5 π ) + 3 ( 6 5 π ) = 11 ( 2 1 ) + 6 15 π = 2 11 + 2 5 π
Evaluating at Lower Limit Then, we evaluate the antiderivative at the lower limit of integration, 2 π :
11 sin ( 2 π ) + 3 ( 2 π ) = 11 ( 1 ) + 2 3 π = 11 + 2 3 π
Subtracting and Simplifying Now, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: ( 2 11 + 2 5 π ) − ( 11 + 2 3 π ) = 2 11 + 2 5 π − 11 − 2 3 π = 2 11 − 2 22 + 2 2 π = − 2 11 + π So the value of the definite integral is − 2 11 + π .
Final Answer Therefore, the exact value of the definite integral is π − 2 11 .
Examples
Definite integrals are used to calculate the area under a curve, which has applications in physics (e.g., finding displacement from a velocity function), engineering (e.g., calculating work done by a force), and economics (e.g., determining consumer surplus). For instance, if you have a function that describes the rate of water flow into a tank, integrating that function over a specific time interval will tell you the total amount of water that entered the tank during that time.
