Find the area of the region under the graph of fx ln x from

Find the area of the region under the graph of f(x) = ln x from x = 1 to x = 5. The velocity of a + sec after leaving the starting line is 100 te^-0.2t F+/sec. What is the distance covered by the in the first 10 sec of its run?

Solution

Solution:

A)

Here;

integral of 1 to 5 log(x) dx = 5 log(5) - 4 = 4.04719

Concept used:

Take the integral:

integral of log(x) dx

For the integrand log(x), integrate by using parts,

integral of f dg = f g- integral g df,

where \ f = log(x), dg = dx,\ df = 1\\/x dx, g = x:\ = x log(x)- integral 1 dx\ The integral of 1 is x:

Answer: => x log(x)-x+constant

B)

As we know;

v(t) = Dx [ S(t) ] where S(t) is the displacement.

Thus what you have is ? 10 0 v(t) dt
? 10 0 100t * e^(-0.2t) dt = 100 * ? 10 0 t * e^(-0.2t) dt (? 10 0 is 0 to 16 integration)

Integration by Parts:
Let u = t, du = dt
dv = e^(-0.2t) dt, v = -5e^(-0.2t)

? t * e^(-0.2t) dt = uv - ? v du
= -5te^(-0.2t) - ? -5e^(-0.2t) dt
= -5te^(-0.2t) - ? 25e^(-0.2t) (-0.2)dt
= -5te^(-0.2t) + 25e^(-0.2t)
= e^(-0.2t) * (-5t - 25)

Thus, the whole expression is S = ? 10 0 v(t) dt

=> 100 * e^(-0.2t) * (-5t - 25) | b=10 a=0

=> 100*(25 - (75/e²) = 1484.98537573