# If # int_0^3 f(x) dx = 8 # then calculate? (A) (i) #int_0^3 2f(x) dx#, (ii) #int_0^3 f(x) + 2 dx# (B) #c# and #d# so that #int_c^d f(x-2) dx #

A) (i)

#int_0^3 \ 2f(x) \ dx = 16 #

A) (ii)#int_0^3 \ f(x) + 2 \ dx = 14 # B)

# c=2 # ;#d = 5 #

We are given that:

Part (A)

Part (B)

We are given that:

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(A) (i) ( \int_0^3 2f(x) , dx = 2 \int_0^3 f(x) , dx = 2 \times 8 = 16 ) (ii) ( \int_0^3 f(x) + 2 , dx = \int_0^3 f(x) , dx + \int_0^3 2 , dx = 8 + 2 \times 3 = 14 )

(B) Let's substitute ( u = x - 2 ), then ( du = dx ). ( \int_c^d f(x-2) , dx = \int_{c-2}^{d-2} f(u) , du ) Since ( \int_0^3 f(x) , dx = 8 ), this means ( \int_0^3 f(u) , du = 8 ). So, ( c-2 = 0 ) and ( d-2 = 3 ). Therefore, ( c = 2 ) and ( d = 5 ).

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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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