Question

Use integration by parts to evaluate the definite integral: ∫ from 4.75 π to − 6 π of 9 t sin ( 6 t ) d t where b=4.75 pi and a=-6pi

Answer 1

To evaluate the definite integral, follow these steps using integration by parts to find the antiderivative of the function and evaluate it using the limits of integration.

Step-by-step explanation:

To evaluate the definite integral ∫ from 4.75π to −6π of 9t sin(6t) dt using integration by parts, we need to follow these steps:

1. Choose u and dv:

• Let u = 9t
• Let dv = sin(6t) dt

Compute du and v:

• Compute du/dt by taking the derivative of u with respect to t: du/dt = 9
• Integrate dv to find v: ∫sin(6t) dt = -cos(6t)/6

Apply the formula for integration by parts:

• Using the formula ∫u dv = uv - ∫v du, we have:
• ∫9t sin(6t) dt = (9t)(-cos(6t)/6) - ∫(-cos(6t)/6)(9) dt

Simplify the integral:

• Simplify the terms in the integral and evaluate it:
• (9t)(-cos(6t)/6) - (9/6) ∫cos(6t) dt = -(3t/2)cos(6t) + (3/2)sin(6t)

Evaluate the definite integral using the limits of integration:

• Substitute the upper limit of integration into the expression and subtract the result of substituting the lower limit:
• (-(3(4.75π)/2)cos(6(4.75π)) + (3/2)sin(6(4.75π))) - (-(3(-6π)/2)cos(6(-6π)) + (3/2)sin(6(-6π)))

After simplifying and evaluating, the definite integral is equal to a specific value.