
Dayv O. answered 05/24/23
Caring Super Enthusiastic Knowledgeable Calculus Tutor
first one
h(x)=d[∫(2tt+4t-6)dt t from 2 to x]/dx
is easy application of FTCpart1
h(x)=2xx-4x-6
second one is integral t from -1 to u(x)=x3
h(x)=(du/dx)d[∫(t)dt t from -1 to u]/du
h(x)=(3x2)u=3x5