∫ (from 0 to log ( 1+ sqrt{ 2) of )}{left(e^x-e^x/2right)}^3{left(e^x-e^x/2right)}^11 wrt x eching

$$\int_{ 0 }^{ \log ( 1+ \sqrt{ 2 } ) } { \left( \frac{ { e }^{ x } - { e }^{ x } }{ 2 } \right) }^{ 3 } { \left( \frac{ { e }^{ x } - { e }^{ x } }{ 2 } \right) }^{ 11 } d x$$

Baholash

$$0$$

Yechim qadamlari

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 3 va 11 ni qo‘shib, 14 ni oling.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}\left(\frac{e^{x}-e^{x}}{2}\right)^{14}\mathrm{d}x$$

0 ni olish uchun e^{x} va -e^{x} ni birlashtirish.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}\left(\frac{0}{2}\right)^{14}\mathrm{d}x$$

Nol bo‘lmagan har qanday sonni nolga ko‘paytirsangiz, nol bo‘ladi.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}0^{14}\mathrm{d}x$$

14 daraja ko‘rsatkichini 0 ga hisoblang va 0 ni qiymatni oling.

$$\int _{0}^{\log_{10}\left(1+\sqrt{2}\right)}0\mathrm{d}x$$

Avval noaniq integralni baholang.

$$\int 0\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, 0 integralini toping.

$$0$$

Xos integral bu integral hisoblashning yuqori chegarasida hisoblangan ifodaning boshlangʻich holatidan chiqarib tashlagan holda integral hisoblashning quyi chegarasida hisoblangan ifodaning boshlangʻich holatidir.

$$0+0$$

Qisqartirish.

$$0$$