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.
To calculate these integrals, we can use a technique called variable substitution. We substitute the given expression in the integrals with a new variable, perform the integration using the new variable, and then revert back to the original variable.
Let's start with the first integral:
A. int0->0.25 f(4t)dt
To find the new variable, let's set 4t equal to the new variable, say u. So, we have:
u = 4t
To get dt in terms of du, we can differentiate both sides with respect to t:
du/dt = 4
Now, solve the above equation for dt:
dt = (1/4) du
Substitute dt and u into the integral:
int0->0.25 f(4t)dt = int0->0.25 f(u) * (1/4) du
Now, evaluate the limits of integration using the substitution:
When t = 0, u = 4(0) = 0
When t = 0.25, u = 4(0.25) = 1
Therefore, the new limits of integration are from 0 to 1. The integral becomes:
int0->0.25 f(4t)dt = (1/4) int0->1 f(u) du
Since Int 0->1 f(t)dt = 11, the integral becomes:
(1/4) * 11 = 11/4 = 2.75
So, A. int0->0.25 f(4t)dt = 2.75.
Similarly, we can proceed with the other integrals:
B. int0->0.25 f(1−4t)dt
Set 1-4t equal to a new variable v:
v = 1 - 4t
Differentiating both sides with respect to t:
dv/dt = -4
dt = - (1/4) dv
Substitute dt and u into the integral:
int0->0.25 f(1−4t)dt = int0->0.25 f(v) * (-1/4) dv
Since Int 0->1 f(t)dt = 11, the integral becomes:
(-1/4) * 11 = -11/4 = -2.75
So, B. int0->0.25 f(1−4t)dt = -2.75.
C. int0.25->0.375 f(3−8t)dt
Set 3-8t equal to a new variable w:
w = 3 - 8t
Differentiating both sides with respect to t:
dw/dt = -8
dt = - (1/8) dw
Substitute dt and u into the integral:
int0.25->0.375 f(3−8t)dt = int0.25->0.375 f(w) * (-1/8) dw
The limits of integration remain the same.
Therefore, using the same logic as above:
C. int0.25->0.375 f(3−8t)dt = - (1/8) * 11 = -11/8 = -1.375
So, C. int0.25->0.375 f(3−8t)dt = -1.375.
I hope this explanation helps you understand how to calculate these integrals using the technique of variable substitution.