nice-calculus-I-calculate-0-1-e-1-log-1-ex-x-e-x-2-dx-

Question Number 128851 by mnjuly1970 last updated on 10/Jan/21

\dots n i c e c a l c u l u s (I) \dots

c a l c u l a t e ::

Ψ = \int_{0}^{1} {(e - 1) \sqrt{l o g (1 + e x - x)} + e^{x^{2}}} d x = ?

Commented by mindispower last updated on 10/Jan/21

{.} f r a c t i o n p a r t ?

Commented by mnjuly1970 last updated on 11/Jan/21

n o m r p o w e r

Answered by Olaf last updated on 11/Jan/21

Let f (x) = (e - 1) \sqrt{\ln (1 + (e - 1) x} + e^{x^{2}}

f is clearly an increasing function and

f (0) = 1, f (1) = e

\Rightarrow \exists a \in] 0, 1 [∖ f (a) = 2

Ψ = \int_{0}^{1} {f (x)} d x = \int_{0}^{a} {f (x)} d x + \int_{a}^{1} {f (x)} d x

Ψ = \int_{0}^{a} (f (x) - 1) d x + \int_{a}^{1} (f (x) - 2) d x

Ψ = \int_{0}^{1} f (x) d x - a - 2 (1 - a)

Ψ = \int_{0}^{1} f (x) d x + a - 2

Let u = \ln (1 + (e - 1) x)

e^{u} = 1 + (e - 1) x

e^{u} d u = (e - 1) d x

I = (e - 1) \int_{0}^{1} \sqrt{\ln (1 + (e - 1) x)} d x

I = (e - 1) \int_{0}^{1} \sqrt{u} \frac{e^{u} d u}{e - 1}

I = \int_{0}^{1} \sqrt{u} e^{u} d u

I = {[\sqrt{x} e^{x} - \frac{\sqrt{π}}{2} erfi (\sqrt{x})]}_{0}^{1} = e - \frac{\sqrt{π}}{2} erfi (1)

J = \int_{0}^{1} e^{x^{2}} d x = {[\frac{\sqrt{π}}{2} erfi (x)]}_{0}^{1} = \frac{\sqrt{π}}{2} erfi (1)

Ψ = I + J + a - 2 = e + a - 2

(a \approx 0, 542)

Commented by mnjuly1970 last updated on 11/Jan/21

g r a t e f u l f o r y o u r w o r k m r o l a f . .

Answered by mnjuly1970 last updated on 11/Jan/21

n o t e : \int_{a}^{b} f (x) d x + \int_{c}^{d} f^{- 1} (x) d x = b d - a c

Ψ = \int_{0}^{1} (e - 1) \sqrt{l o g (1 + x (e - 1)} d x

+ \int_{0}^{1} e^{x^{2}} d x

∴ Ψ \overset{[u = (1 + x (e - 1)]}{=} \int_{1}^{e} \sqrt{l o g (u)} d u + \int_{0}^{1} e^{x^{2}} d x

= \int_{1}^{e} \sqrt{l o g (x)} + \int_{0}^{1} e^{x^{2}} d x = e

n o t e : f (x) = \sqrt{l o g (x)} \Leftrightarrow f^{- 1} (x) = e^{x^{2}}

Answered by mnjuly1970 last updated on 11/Jan/21