Question

@perl

Answer 1

we know that
$$

\Large {
\int_{1}^{6} g(x)~dx = \int_{1}^{4}g(x)~dx + \int_{4}^{6}g(x)~dx
\\~\\ 10 = \int_{1}^{4}g(x) ~dx + (-2)
\\ \therefore \\
\\10 + 2 = \int_{1}^{4}g(x)


}
$$

Answer 2

but be careful, the question is asking for integral 2g(x) , not g(x)

Answer 3

so the value would be 12?

Answer 4

not quite

Answer 5

$$
\Large {
\int_{1}^{6} g(x)~dx = \int_{1}^{4}g(x)~dx + \int_{4}^{6}g(x)~dx
\\~\\ 10 = \int_{1}^{4}g(x) ~dx + (-2)
\\ \therefore \\
\\10 + 2 = \int_{1}^{4}g(x)
\\ \therefore \\
\\12 = \int_{1}^{4}g(x)
\\ \therefore \\
\\2\cdot 12 = 2\cdot \int_{1}^{4}g(x) = \int_{1}^{4}2\cdot g(x)

}

$$

Answer 6

ohhh 24

Answer 7

thank you!!

Answer 8

I got $$ \Large 4e^{x^3} $$

Answer 9

yay, thanks(: