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(06.04 MC)

If
\int\limits^3_ {-2} \, [2f(x)+2]dx=18 and
\int\limits^1_ {-2} \, f(x)dx =8, then
\int\limits^3_ {1} \, f(x)dx is equal to which of the following?

4

0

−2

−4

User Everett
by
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1 Answer

23 votes
23 votes


\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{\frac{2 { \int}^(3) _(2) }{2} = (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}

User Yogesh Wadhwa
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