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 #

Question

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 #

Samantha Adkison · Accepted Answer

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: # 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 #

Evelyn Agard · Answer

#(a)(i): 16; (ii): 14; (b): c=2, 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.!

Charles Arocha · Answer

undefined (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 $.

HIX Tutor · Answer

undefined Sure! Let's solve it step-by-step.

**(A)**

1. **(i)** Calculate ∫₀³ 2f(x) dx.
   - Use the rule: ∫a^b k*f(x) dx = k * ∫a^b f(x) dx where k is a constant.
   - So, here: ∫₀³ 2f(x) dx = 2 * ∫₀³ f(x) dx.
   - We know ∫₀³ f(x) dx = 8.
   - Therefore, ∫₀³ 2f(x) dx = 2 * 8 = 16.

2. **(ii)** Calculate ∫₀³ (f(x) + 2) dx.
   - Use the rule: ∫a^b (f(x) + g(x)) dx = ∫a^b f(x) dx + ∫a^b g(x) dx.
   - Here, g(x) = 2.
   - So, ∫₀³ (f(x) + 2) dx = ∫₀³ f(x) dx + ∫₀³ 2 dx.
   - We already found ∫₀³ f(x) dx = 8.
   - Now, calculate ∫₀³ 2 dx: 
     - This is 2 * (3 - 0) = 2 * 3 = 6.
   - So, ∫₀³ (f(x) + 2) dx = 8 + 6 = 14.

**(B)**

To find c and d for ∫c^d f(x-2) dx:

1. Use a substitution:
   - Let u = x - 2.
   - Then, x = u + 2.
   - When x = 0, u = 0 - 2 = -2.
   - When x = 3, u = 3 - 2 = 1.
2. So, c = -2 and d = 1.

**Final Answers:**
(A) (i) 16, (ii) 14  
(B) c = -2, d = 1.

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 #