Question

(06.04 MC)

Answer 1

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

◉ $\large\bm{ -4}$

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$