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 #

We are given that:

# int_0^3 \ f(x) \ dx = 8 #

Part (A)

(i) #int_0^3 \ 2f(x) \ dx = 2 \ int_0^3 \ f(x) \ dx #
# " " = 2* 8 #
# " " = 16 #

(ii) #int_0^3 \ f(x) + 2 \ dx = int_0^3 \ f(x) \ dx + int_0^3 \ 2 \ dx #
# " " = int_0^3 \ f(x) \ dx + 2 \ int_0^3 \ dx #
# " " = 8 + 2[x]_0^3 #
# " " = 8 + 2(3-0) #
# " " = 8 + 6 #
# " " = 14 #

Part (B)

We are given that:

#int_c^d \ f(x-2) \ dx = 8 #

The graph of #y=f(x-2)# represents a translation of #y=f(x)# by a shift to the right by two units.

Thus, as we know that # int_0^3 \ f(x) \ dx = 8 # then

# c-2 = 0 => c=2 #
# d-2 = 3 => d = 5 #

Given that, for a fun. #f, int_0^3 f(x)dx=8.#

#(a)(i): int_0^3{2f(x)}dx=2int_0^3f(x)dx=2*8=16.#

#(a)(ii): int_0^3{f(x)+2}dx=int_0^3f(x)dx+int_0^3 2dx,#

#=8+2int_0^3 1dx,#

#=8+2[x]_0^3,#

#=8+2[3-0],#

#=8+6,#

#=14.#

#(b): int_c^df(x-2)dx=8.#

Let, #(x-2)=t," so that, "dx=dt.#

Also, when, #x=c, t=x-2=c-2.#

Similarly, when, #x=d, t=d-2.#

#:. int_c^d f(x-2)dx=int_(c-2)^(d-2)f(t)dt.#

But, #int_c^df(x-2)dx=8=int_0^3 f(x)dx=8=int_0^3f(t)dt.#

#:. int_(c-2)^(d-2)f(t)dt=8=int_0^3f(t)dt.#

Evidently, #c-2=0 rArr c=2, and, d-2=3 rArr d=5.#

#;. (c,d)=(2,5).#

Enjoy Maths.!