Calculus
posted by Cady .
Please help me calculate these integrals.
Suppose that Int 0>1 f(t)dt=11 . Calculate each of the following.
A. int0>0.25 f(4t)dt=
B. int0>0.25 f(1−4t)dt=
C. int0.25>0.375 f(3−8t)dt=

Try substitution:
u=4t
du=4dt, or
dt=(1/4)du
then
∫f(4t)dt [0,0.25]
=∫f(u)(1/4)du [0,4*0.25]
=(1/4)∫f(u)du [0,1]
=(1/4)*11
=11/4
The other problems should work similarly.
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