Given the inequality e^x>x+1, deduct the inequality 4/3=

We'll take the given constraint and we'll create a
function:

h(x) = e^x - x - 1 >
0

We'll differentiate the function with respect to
x:

h'(x) = e^x - 1

If x belongs to the
interval (-infinite ; 0], the function h(x) is decreasing.

If x
belongs to the interval (0 ;infinite), the function h(x) is
increasing.

So, f(x) > = 0 =
h(0)

Since e^x > x + 1 => e^(x^2) > x^2 +
1

We'll integrate over the closed interval
[0;1].

Int e^(x^2) dx> Int (x^2 + 1)
dx

Int (x^2 + 1) dx = x^3/3 + x

We'll
apply Leibniz Newton:

Int (x^2 + 1) dx = F(1) -
F(0)

F(1) - F(0) = 1/3 + 1 - 0/3 -
0

F(1) - F(0) = 4/3

Int (x^2 + 1) dx =
4/3

Since Int e^(x^2) dx> Int (x^2 + 1) dx =
4/3, the given inequality is verified.