Given the inequality e^x>x+1, deduct the inequality 4/3=<Integral of e^(x^2), if the limits of integration are x=0 and x=1.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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