If # int_0^1 f(t) dt = 19# then find?: (A) #int_0^0.125 f(8t) dt#, (B) #int_0^0.25 f(1−4t) dt#, and (C) #int_0.4^0.5 f(5-10t) dt#

Answer 1

# int_0^0.125 \ f(8t) \ dt = 19/8#

# int_0^0.25 \ f(1−4t) \ dt = 19/4#

# int_0.4^0.5 \ f(5-10t) \ dt = 9/10#

We have:

# int_0^1 \ f(t) \ dt = 19#
We first note that the above result is independent of the variable #t#, thus:
# int_0^1 \ f(t) \ dt = int_0^1 \ f(u) \ du = 19 #

Part A

Find # I_1 = int_0^0.125 \ f(8t) \ dt#

We can perform a substitution, Let:

#u = 8t => (du)/(dt) = 8 => 1/8(du)/(dt) = 1#

And we change the limits of integration:

# t={ (0), (0.125) :} => u={ (0), (1) :}#

If we substitute into the integral then we get:

# I_1 = int_0^1 \ f(u) \ (1/8) \ du# # \ \ \ = 1/8 \ int_0^1 \ f(u) \ du# # \ \ \ = 1/8 * 19# # \ \ \ = 19/8#

Part B

Find # I_2 = int_0^0.25 \ f(1−4t) \ dt #

We can perform a substitution, Let:

#u = 1-4t => (du)/(dt) = -4 => -1/4(du)/(dt) = 1#

And we change the limits of integration:

# t={ (0), (0.25) :} => u={ (1), (0) :}#

If we substitute into the integral then we get:

# I_2 = int_1^0 \ f(u) \ (-1/4) \ du# # \ \ \ = -1/4 \ int_1^0 \ f(u) \du# # \ \ \ = 1/4 \ int_0^1 \ f(u) \du# # \ \ \ = 1/4 * 19# # \ \ \ = 19/4#

Part C

Find # I_3 = int_0.4^0.5 \ f(5-10t) \ dt#

We can perform a substitution, Let:

#u = 5-10t => (du)/(dt) = -10 => -1/10(du)/(dt) = 1#

And we change the limits of integration:

# t={ (0.4), (0.5) :} => u={ (1), (0) :}#

If we substitute into the integral then we get:

# I_3 = int_1^0 \ f(u) \ (-1/10) \ du# # \ \ \ = -1/10 \ int_1^0 \ f(u) du# # \ \ \ = 1/10 \ int_0^1 \ f(u) du# # \ \ \ = 1/10 * 19# # \ \ \ = 19/10#
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Answer 2

Let ( F(x) ) be the antiderivative of ( f(x) ). Given that ( \int_{0}^{1} f(t) , dt = 19 ), we can use the properties of definite integrals to solve for the required integrals.

(A) ( \int_{0}^{0.125} f(8t) , dt ): Let ( u = 8t ), then ( du = 8 , dt ) or ( dt = \frac{1}{8} du ). Substitute limits and the function: ( \int_{0}^{0.125} f(8t) , dt = \int_{0}^{1} f(u) \cdot \frac{1}{8} , du = \frac{1}{8} \int_{0}^{1} f(u) , du ). Since ( \int_{0}^{1} f(t) , dt = 19 ), we have ( \frac{1}{8} \times 19 = \frac{19}{8} ).

(B) ( \int_{0}^{0.25} f(1-4t) , dt ): Let ( v = 1-4t ), then ( dv = -4 , dt ) or ( dt = -\frac{1}{4} dv ). Substitute limits and the function: ( \int_{0}^{0.25} f(1-4t) , dt = \int_{1}^{0} f(v) \cdot (-\frac{1}{4}) , dv = -\frac{1}{4} \int_{0}^{1} f(v) , dv ). Since ( \int_{0}^{1} f(t) , dt = 19 ), we have ( -\frac{1}{4} \times 19 = -\frac{19}{4} ).

(C) ( \int_{0.4}^{0.5} f(5-10t) , dt ): Let ( w = 5-10t ), then ( dw = -10 , dt ) or ( dt = -\frac{1}{10} dw ). Substitute limits and the function: ( \int_{0.4}^{0.5} f(5-10t) , dt = \int_{5}^{4} f(w) \cdot (-\frac{1}{10}) , dw = -\frac{1}{10} \int_{4}^{5} f(w) , dw ). Since ( \int_{0}^{1} f(t) , dt = 19 ), we have ( -\frac{1}{10} \times 19 = -\frac{19}{10} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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