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

Commented by mindispower last updated on 10/Jan/21

Commented by mnjuly1970 last updated on 11/Jan/21

Answered by Olaf last updated on 11/Jan/21
![Let f(x) = (e−1)(√(ln(1+(e−1)x))+e^x^2 f is clearly an increasing function and f(0) = 1, f(1) = e ⇒ ∃a∈]0,1[ \ f(a) = 2 Ψ = ∫_0 ^1 {f(x)}dx = ∫_0 ^a {f(x)}dx+∫_a ^1 {f(x)}dx Ψ = ∫_0 ^a (f(x)−1)dx+∫_a ^1 (f(x)−2)dx Ψ = ∫_0 ^1 f(x)dx−a−2(1−a) Ψ = ∫_0 ^1 f(x)dx+a−2 Let u = ln(1+(e−1)x) e^u = 1+(e−1)x e^u du = (e−1)dx I = (e−1)∫_0 ^1 (√(ln(1+(e−1)x)))dx I = (e−1)∫_0 ^1 (√u)((e^u du)/(e−1)) I = ∫_0 ^1 (√u)e^u du I = [(√x)e^x −((√π)/2)erfi((√x))]_0 ^1 = e−((√π)/2)erfi(1) J = ∫_0 ^1 e^x^2 dx = [((√π)/2)erfi(x)]_0 ^1 = ((√π)/2)erfi(1) Ψ = I+J+a−2 = e+a−2 (a ≈ 0,542)](https://www.tinkutara.com/question/Q128875.png)
Commented by mnjuly1970 last updated on 11/Jan/21

Answered by mnjuly1970 last updated on 11/Jan/21
![note: ∫_(a ) ^( b) f(x)dx+∫_c ^( d) f^( −1) (x)dx=bd−ac Ψ=∫_0 ^( 1) (e−1)(√(log(1+x(e−1))) dx + ∫_0 ^( 1) e^x^2 dx ∴ Ψ=^([u=(1+x(e−1)]) ∫_1 ^( e) (√(log(u))) du+∫_0 ^( 1) e^x^2 dx =∫_1 ^( e) (√(log(x))) +∫_0 ^( 1) e^x^2 dx=e note: f(x)=(√(log(x))) ⇔ f^( −1) (x)=e^x^2](https://www.tinkutara.com/question/Q128909.png)
Answered by mnjuly1970 last updated on 11/Jan/21
