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PSYCHO15rus [73]
2 years ago
12

(06.04 MC)

Mathematics
1 answer:
Andru [333]2 years ago
7 0

\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}

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