Question

Evaluate the integral. ∫0π/2​[5cos17ti−sin7tj+sin24tk]dt ∫0π/2​[5cos17ti−sin7tj+sin24tk]dt=()i+(∣)j+∣k (Type exact answers, using π as needed.)

Answer 1

To evaluate the given integral, we can split it into three separate integrals for each component of the vector. The integrals can then be evaluated individually using the sine and cosine functions. Finally, the results can be summed up to obtain the final value of the integral.

Step-by-step explanation:

To evaluate the integral ∫[5cos(17t)i - sin(7t)j + sin(24t)k]dt over the interval [0, π/2], we can use the linearity property of integration to evaluate each component separately.

We have:

∫5cos(17t)i dt = 5∫cos(17t) dt = (5/17)sin(17t)i + C

∫-sin(7t)j dt = -∫sin(7t) dt = -(1/7)cos(7t)j + C

∫sin(24t)k dt = -(1/24)cos(24t)k + C

Thus, the definite integral evaluated from 0 to π/2 is (5/17)sin(17t) + -(1/7)cos(7t) + -(1/24)cos(24t) evaluated from 0 to π/2.