Question

高数题， 谢谢！

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Answer 1

1.
s=∫20t(5t²-3)³dt
=2∫(5t²-3)³d(5t²-3)
=2·¼·(5t²-3)⁴+C
=½(5t²-3)⁴+C
t=1，s(1)=12代入
½(5t²-3)⁴+C=12
C=4
s=½(5t²-3)⁴+4
2.
∫[0：π/2]2sin²xcosxdx
=2∫[0：π/2]sin²xd(sinx)
=2·⅓sin³x|[0，π/2]
=⅔[sin³(π/2)-sin³0]
=⅔(1-0)
=⅔
3.
∫[0：π/6](1+sin3t)cos3t dt
=⅓∫[0：π/6](1+sin3t)d(1+sin3t)
=⅓·½(1+sin3t)²|[0：π/6]
=(1/6)[(1+sinπ/2)² -(1+sin0)²]
=(1/6)[(1+1)²-(1+0)²]
=(1/6)(4-1)
=½