Question

Please help! Will FAN AND MEDAL!

Suppose the integral from 2 to 6 of g of x, dx equals 12 and the integral from 5 to 6 of g of x, dx equals negative 3, find the value of the integral from 2 to 5 of 3 times g of x, dx .

a. 45
b. 15
c. 5
d. 3

Answer 1

\( \int_2^6 g(x) dx = \int_2^5 g(x)dx+ \int_5^6 g(x)dx \)
or, in a more synthetic form: \(\int_a^b = \int_a^c + \int_c^b\). Use this to compute \(\int_2^5g(x)dx\) first.

Answer 2

I don't get it?

Answer 3

so `the integral from 2 to 6 of g of x, dx equals 12 and the integral from 5 to 6 of g of x, dx equals negative 3` transform
\[\int_2^6 g(x) dx = \int_2^5 g(x)dx+ \int_5^6 g(x)dx \qquad(1)\]
into
\[ 12 = \int_2^5 g(x)dx + (-3) \qquad (2)\]

Answer 4

my bad. I am sorry.

Answer 5

So that \[\int_2^5 g(x)dx = 12+3 = 15.\]
Now can you compute
\[ \int_2^5 3g(x)dx \quad?\]

Answer 6

I replace g(x) with 15 right? but do I do it like this 3(15)?

Answer 7

Not that quickly. you don't know the value of \(g(x)\), only the value of the integral between 2 and 5. Instead use:
\[\int_2^5 \alpha g(x) dx = \alpha \int_2^5 g(x)dx\]

Answer 8

okay so now I replace g(x) with 15?

Answer 9

No, 15 is the value of the integral:
\[ \int_2^5 3g(x)dx = 3 \underbrace{\int_2^5 g(x)dx}_{=15} = ?\]

Answer 10

\[\frac{ 63g }{ 2 }\]
Okay I get it, this is what I got when I plugged it into the calculator

Answer 11

I do \(3\times 15 = 45\). We don't know \(g(x)\) and we don't need to.

Answer 12

Why don't we need to know g(x)

Answer 13

the situation is as in the sketch. there's a function \(g\) there and all we know is the value of the area under the curve.
total area: 12
last part: -3
=> first part: 15