calculate-f-x-2-x-dx-real-

Question Number 74888 by abdomathmax last updated on 03/Dec/19

c a l c u l a t e f (α) = \int \sqrt{x^{2} - x + α} d x (α r e a l)

Commented by MJS last updated on 03/Dec/19

no borders ?

Commented by mathmax by abdo last updated on 03/Dec/19

y e s

Commented by mathmax by abdo last updated on 03/Dec/19

f (α) = \int \sqrt{x^{2} - x + α} d x

l e t s o l v e x^{2} - x + α = 0 \to Δ = 1 - 4 α

c a s e (1) 1 - 4 α < 0 \Rightarrow x^{2} - x + α = x^{2} - 2 \frac{x}{2} + \frac{1}{4} + α - \frac{1}{4}

= {(x - \frac{1}{2})}^{2} + \frac{4 α - 1}{4} (4 α - 1 > 0) \Rightarrow

f (α) =_{x - \frac{1}{2} = \frac{\sqrt{4 α - 1}}{2} s h (u)} \int \sqrt{\frac{4 α - 1}{4}} c h (u) \frac{\sqrt{4 α - 1}}{2} c h (u) d u

= \frac{4 α - 1}{4} \int {c h}^{2} (u) d u = \frac{4 α - 1}{4} \int \frac{1 + c h (2 u)}{2} d u

= \frac{(4 α - 1)}{8} u + \frac{4 α - 1}{16} s h (2 u) + C

= \frac{(4 α - 1)}{8} a r g s h (\frac{2 x - 1}{\sqrt{4 α - 1}}) + \frac{4 α - 1}{8} s h (u) c h (u) + c

= \frac{(4 α - 1)}{8} l n (\frac{2 x - 1}{\sqrt{4 α - 1}} + \sqrt{1 + {(\frac{2 x - 1}{\sqrt{4 α - 1}})}^{2}}) + \frac{4 α - 1}{8} \times \frac{2 x - 1}{\sqrt{4 α - 1}} \sqrt{1 + {(\frac{2 x - 1}{\sqrt{4 α - 1}})}^{2}} + C

c a s e (2) 1 - 4 α > 0 \Rightarrow x_{1} = \frac{1 + \sqrt{1 - 4 α}}{2} a n d x_{2} = \frac{1 - \sqrt{1 - 4 α}}{2}

f (α) = \int \sqrt{(x - x_{1}) (x - x_{2})} d x c h a n g e m e n t \sqrt{x - x_{1}} = u g i v e x - x_{1} = u^{2}

f (α) = \int u \sqrt{u^{2} + x_{1} - x_{2}} (2 u) d u

= 2 \int u^{2} \sqrt{u^{2} + \sqrt{1 - 4 α}} d u w e d o t h e c h a n g e m e n t u = (^{4} \sqrt{1 - 4 α}) s h (z)

\Rightarrow f (α) = 2 \int \sqrt{1 - 4 α} \sqrt{1 - 4 α} {s h}^{2} (z) c h (z) (^{4} \sqrt{1 - 4 α}) c h (z) d z

= 2 (1 - 4 α) (^{4} \sqrt{1 - 4 α}) \int {s h}^{2} (z) {c h}^{2} (z) d z

= \frac{1}{2} {(1 - 4 α)}^{\frac{5}{4}} \int {s h}^{2} (2 z) d z = \dots . b e c o n t i n u e d \dots

Answered by MJS last updated on 03/Dec/19

\int \sqrt{x^{2} - x + α} d x = \int \sqrt{{(x - \frac{1}{2})}^{2} + α - \frac{1}{4}} d x =

[t = \frac{2 x - 1}{\sqrt{4 α - 1}} \to d x = \frac{\sqrt{4 α - 1}}{2} d t]

= \frac{4 α - 1}{4} \int \sqrt{t^{2} + 1} d t =

= \frac{4 α - 1}{8} (t \sqrt{t^{2} + 1} + \ln (t + \sqrt{t^{2} + 1}) =

\dots

= \frac{(2 x - 1) \sqrt{x^{2} - x + α}}{4} + \frac{4 α - 1}{8} \ln (2 x - 1 + 2 \sqrt{x^{2} - x + α}) + C

Facebook Tweet Pin