∫ (sqrt{x}-5)(sqrt{x}+5) wrt x= eching

$$\int ( \sqrt { x } - 5 ) ( \sqrt { x } + 5 ) d x =$$

Baholash

$$\frac{x^{2}}{2}-25x+С$$

Yechim qadamlari

Hisoblang: \left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\int \left(\sqrt{x}\right)^{2}-5^{2}\mathrm{d}x$$

2 daraja ko‘rsatkichini \sqrt{x} ga hisoblang va x ni qiymatni oling.

$$\int x-5^{2}\mathrm{d}x$$

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$\int x-25\mathrm{d}x$$

Summani muddatma-muddat integratsiya qiling.

$$\int x\mathrm{d}x+\int -25\mathrm{d}x$$

k\neq -1 uchun integral \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} boʻlgani uchun, \int x\mathrm{d}x integralni \frac{x^{2}}{2} bilan almashtiring.

$$\frac{x^{2}}{2}+\int -25\mathrm{d}x$$

\int a\mathrm{d}x=ax umumiy integrallar qoidasi jadvalidan foydalanib, -25 integralini toping.

$$\frac{x^{2}}{2}-25x$$

Агар F\left(x\right)f\left(x\right) ning dastlabki holati boʻlsa, u holatda f\left(x\right) ning barcha dastlabki holatlari toʻplami F\left(x\right)+C tarafidan belgilanadi. Shu sababli natijaga C\in \mathrm{R} integrallash konstantasini qoʻshing.

$$\frac{x^{2}}{2}-25x+С$$

x ga nisbatan hosilani topish

$$x-25$$