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Consider two functions f and g on [1, 8] such that integral^8_1 f(x) dx = 9, integral^8_1 g(x) dx = 5, integral^8_5 f(x) dx = 4, and integral^5_1 g (x) dx = 3. Evaluate the following integrals. a. integral^5_1 2f(x) dx = (Simplify your answer.) b. integral^8_1 (f(x) - g (x)) dx = (Simplify your answer.) c. integral^5_1 (f (x) - g (x)) dx = (Simplify your answer.) d. integral^8_5 (g(x) - f(x)) dx = (Simplify your answer.) e. integral^8_5 7g(x) dx = (Simplify your answer.) f. integral^1_5 3f(x) dx = (Simplify your answer.)

Respuesta :

LammettHash

I'll abbreviate the definite integral with the notation,

[tex]I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx[/tex]

We're given

  • [tex]I(f,1,8)=9[/tex]
  • [tex]I(g,1,8)=5[/tex]
  • [tex]I(f,5,8)=4[/tex]
  • [tex]I(g,1,5)=3[/tex]

Recall that the definite integral is additive on the interval [tex][a,b][/tex], meaning for some [tex]c\in[a,b][/tex] we have

[tex]I(f,a,b)=I(f,a,c)+I(f,c,b)[/tex]

The definite integral is also linear in the sense that

[tex]I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)[/tex]

for some constant scalars [tex]k,\ell[/tex].

Also, if [tex]a\ge b[/tex], then

[tex]I(f,a,b)=-I(f,b,a)[/tex]

a. [tex]I(2f,1,5)=2I(f,1,5)=2(I(f,1,8)-I(f,5,8))=2(9-4)=\boxed{10}[/tex]

b. [tex]I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}[/tex]

c. [tex]I(f-g,1,5)=I(f,1,5)-I(g,1,5)=\dfrac{I(2f,1,5)}2-I(g,1,5)=10-3=\boxed{7}[/tex]

d. [tex]I(g-f,5,8)=I(g,5,8)-I(f,5,8)=(I(g,1,8)-I(g,1,5))-I(f,5,8)=(5-3)-4=\boxed{-2}[/tex]

e. [tex]I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}[/tex]

f. [tex]I(3f,5,1)=3I(f,5,1)=-3I(f,1,5)=-\dfrac32I(2f,1,5)=-\dfrac32(10)=\boxed{-15}[/tex]

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