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sveticcg [70]
3 years ago
13

Suppose the integral from 2 to 8 of g of x, dx equals 12, and the integral from 6 to 8 of g of x, dx equals negative 3, find the

value of 2 times the integral from 2 to 6 of g of x, dx .
7.5
15
18
30
Mathematics
1 answer:
Artyom0805 [142]3 years ago
8 0

\displaystyle\int_2^8g(x)\,\mathrm dx=12

\displaystyle\int_6^8g(x)\,\mathrm dx=-3

Use the fact that integrals are additive on their intervals. Mathematically, if c\in[a,b], then

\displaystyle\int_a^bg(x)\,\mathrm dx=\int_a^cg(x)\,\mathrm dx+\int_c^bg(x)\,\mathrm dx

So we have

\displaystyle2\int_2^6g(x)\,\mathrm dx=2\left(\int_2^8g(x)\,\mathrm dx-\int_6^8g(x)\,\mathrm dx\right)=2(12-(-3))=30

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<em>hope </em><em>it </em><em>helps</em><em> </em><em>and </em><em>your </em><em>day </em><em>will </em><em>fine</em>

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