Evaluate the integral. Give result in exact form.
[tex]$\int_{\frac{x}{2}}^{\frac{5}{8}}(3 \cos (x)+6) d x=\square$[/tex]
Answer (2)
Find the antiderivative of the integrand: 3 sin ( x ) + 6 x .
Evaluate the antiderivative at the upper limit: 3 sin ( 8 5 ) + 4 15 .
Evaluate the antiderivative at the lower limit: 3 sin ( 2 x ) + 3 x .
Subtract the value at the lower limit from the value at the upper limit: 3 sin ( 8 5 ) + 4 15 − 3 sin ( 2 x ) − 3 x .
Explanation
Understanding the Problem We are asked to evaluate the definite integral ∫ 2 x 8 5 ( 3 cos ( x ) + 6 ) 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 subtract the results.
Finding the Antiderivative First, let's find the antiderivative of 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 the Upper Limit Now, we need to evaluate the antiderivative at the upper limit of integration, which is 8 5 . So we have 3 sin ( 8 5 ) + 6 ( 8 5 ) = 3 sin ( 8 5 ) + 8 30 = 3 sin ( 8 5 ) + 4 15 .
Evaluating at the Lower Limit Next, we evaluate the antiderivative at the lower limit of integration, which is 2 x . So we have 3 sin ( 2 x ) + 6 ( 2 x ) = 3 sin ( 2 x ) + 3 x .
Subtracting the Values Finally, we subtract the value of the antiderivative at the lower limit from the value at the upper limit: ( 3 sin ( 8 5 ) + 4 15 ) − ( 3 sin ( 2 x ) + 3 x ) = 3 sin ( 8 5 ) + 4 15 − 3 sin ( 2 x ) − 3 x .
Final Result Therefore, the definite integral is 3 sin ( 8 5 ) + 4 15 − 3 sin ( 2 x ) − 3 x .
Examples
Definite integrals are used extensively in physics and engineering to calculate quantities such as the work done by a force, the area under a curve, or the volume of a solid. For example, if you want to calculate the total distance traveled by an object with a given velocity function over a certain time interval, you would use a definite integral. Similarly, in electrical engineering, definite integrals are used to calculate the total charge that flows through a circuit over a period of time.
To evaluate the integral ∫ 2 x 8 5 ( 3 cos ( x ) + 6 ) d x , we find the antiderivative, evaluate it at the upper and lower limits, and then subtract these values. The final result is 3 sin ( 8 5 ) + 4 15 − 3 sin ( 2 x ) − 3 x .
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