Question

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.)

Answer 1

I'll abbreviate the definite integral with the notation,

$I(f(x),a,b)=\int_a^bf(x)\,\mathrm dx$

We're given

• $I(f,1,8)=9$
• $I(g,1,8)=5$
• $I(f,5,8)=4$
• $I(g,1,5)=3$

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

$I(f,a,b)=I(f,a,c)+I(f,c,b)$

The definite integral is also linear in the sense that

$I(kf+\ell g,a,b)=kIf(a,b)+\ell I(g,a,b)$

for some constant scalars $k,\ell$.

Also, if $a\ge b$, then

$I(f,a,b)=-I(f,b,a)$

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

b. $I(f-g,1,8)=I(f,1,8)-I(g,1,8)=9-5=\boxed{4}$

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

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

e. $I(7g,5,8)=7I(g,5,8)=7(5-3)=\boxed{14}$

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