∫ u2du
d(2x–7)=2dx
dx=(1/2)d(2x–7)
=(1/2) ∫ 10(2x–7)2d(2x–7)=((1/2)·(2x–7)3/3)|10=
=(1/6)·(2·1–7)3–(1/6)·(2)·0–7)3=(1/6)(–5)3–(1/6)·(–7)3=
=(1/6)·(73–53)=(1/6)·(7–5)·(72+7·5+52)=109/3= 36 (1/3)
2.
Метод замены
x=t6
√x=t3
∛x=t2
dx=6t5dt
Пределы интегрирования:
если x=0, то t=0
если х=3, то t=31/6
= ∫31/60 (√2·t3+t2)6t5dt=
=6√2 ∫31/60t8dt+6 ∫31/60t7dt=
=6√2·(t9/9)|31/60+6·(t8/8)|31/60=
=(2/3)√2·33/2+(3/4)38/6=
=(2/3)√2·3√3+(3/4)·3∛3=
= 2√6+(9/4)·∛3