Question

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

Answer 1

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

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

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

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