Question

I need help with this problem. I will upload a picture.

Answer 1

(integral)ln(x) dx

set
u = ln(x), dv = dx
then we find
du = (1/x) dx, v = x

substitute

(integral) ln(x) dx = (integral) u dv

and use integration by parts

= uv - (integral) v du

substitute u=ln(x), v=x, and du=(1/x)dx

= ln(x) x - (integral) x (1/x) dx
= ln(x) x - (integral) dx
= ln(x) x - x + C
= x ln(x) - x + C.

Answer 2

where are you getting ln(x) from?

Answer 3

we are given

\[\Large \int_{1}^{10}f(x)dx = 4\]
\[\Large \int_{10}^{3}f(x)dx = 7\]

we want to find

\[\Large \int_{1}^{3}f(x)dx\]

Answer 4

srry this was for someone else

Answer 5

the first step is to flip the limits of integration on

\[\Large \int_{10}^{3}f(x)dx = 7\]

to get

\[\Large \int_{10}^{3}f(x)dx = 7\]

\[\Large -\int_{3}^{10}f(x)dx = 7\]

\[\Large \int_{3}^{10}f(x)dx = -7\]

take note of the sign change

Answer 6

Now let

\[\Large y = \int_{1}^{3}f(x)dx\]

Answer 7

\[\Large \int_{1}^{3}f(x)dx + \int_{3}^{10}f(x)dx = \int_{1}^{10}f(x)dx\]

\[\Large y + (-7) = 4\]

and I'm sure you see how to isolate y at this point

Answer 8

i don't get why the sign changed @jim_thompson5910

Answer 9

the integral from a to b is equal to the exact area under the curve

if the function curve is a velocity function v(t), then the exact area under the curve is equal to the displacement of the object