12 stie careva cum se face?

Răspuns:

Explicație pas cu pas:

[tex]\sqrt{x+6-6\sqrt{x-3}} =\sqrt{x-3-2*3*\sqrt{x-3}+3+6 } =\sqrt{(\sqrt{x-3})^{2} -2*3*\sqrt{x-3}+3^{2}}= \sqrt{(\sqrt{x-3} -3)^{2}} = | \sqrt{x-3} -3 |\\[/tex]

daca √(x-3)-3≥0, adică √(x-3)≥3, x-3≥9, x≥12, atunci |√(x-3)-3|= √(x-3)-3.

dacă x<12, atunci  |√(x-3)-3|=3-√(x-3). Acum vom avea:

[tex]\int\limits^{19}_{3} {\sqrt{x+6-6\sqrt{x-3} } } \, dx =\int\limits^{12}_{3} {(3-\sqrt{x-3})} \, dx+\int\limits^{19}_{12} {(\sqrt{x-3}-3)} \, dx[/tex]

Sper mai departe să te descurci, că nu e atât de complicat...

Succese!

[tex]\displaystyle I = \int_{3}^{19}\sqrt{x+6-6\sqrt{x-3}}\,dx \\ \\\\ \sqrt{x-3} = t \Rightarrow x-3 = t^2 \Rightarrow x = t^2+3 \Rightarrow dx =2t\, dt \\ x = 3 \Rightarrow t = 0 \\ x = 19 \Rightarrow t = 4 \\ \\\\ I = \int_{0}^4\Big(\sqrt{t^2+9 -6t}\Big)\cdot 2t\, dt = \int_{0}^42t\sqrt{t^2-6t+9}\, dt = \\ \\ = \int_{0}^4 2t\sqrt{(t-3)^2}\, dt = \int_{0}^4 2t|t-3|\, dt = \\ \\= \int_{0}^3 2t(3-t)\, dt+\int_{3}^4 2t(t-3)\, dt =[/tex]

[tex]= \Big(3t^2-\dfrac{2t^3}{3}\Big)\Bigg|_{0}^3 +\Big(\dfrac{2t^3}{3}-3t^2\Big)\Bigg|_{3}^4 = \\ \\ = 2\cdot (3\cdot 3^2-2\cdot 3^2)+\dfrac{2\cdot 4^3}{3}-3\cdot 4^2 = \\ \\ = 18+\dfrac{128}{3}-48 = \dfrac{54+128-144}{3} = \boxed{\dfrac{38}{3}}[/tex]

⇒ E) corect