Datrys ∫ (from 4 to 9) of (sqrt{x}+1)sqrt{x wrt x

$$\int _ { 4 } ^ { 9 } ( \sqrt { x } + 1 ) \sqrt { x } d x$$

Enrhifo

$$\frac{271}{6}\approx 45.166666667$$

Camau Ateb

Defnyddio’r briodwedd ddosbarthu i luosi \sqrt{x}+1 â \sqrt{x}.

$$\int _{4}^{9}\left(\sqrt{x}\right)^{2}+\sqrt{x}\mathrm{d}x$$

Cyfrifo \sqrt{x} i bŵer 2 a chael x.

$$\int _{4}^{9}x+\sqrt{x}\mathrm{d}x$$

Gwerthuso’r integryn amhenodol yn gyntaf.

$$\int x+\sqrt{x}\mathrm{d}x$$

Integreiddio'r swm fesul term.

$$\int x\mathrm{d}x+\int \sqrt{x}\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x\mathrm{d}x gyda \frac{x^{2}}{2}.

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

Ailysgrifennwch \sqrt{x} fel x^{\frac{1}{2}}. Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{\frac{1}{2}}\mathrm{d}x gyda \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. Symleiddio.

$$\frac{x^{2}}{2}+\frac{2x^{\frac{3}{2}}}{3}$$

Yr integryn pendant yw integryn amhendant y mynegiant wedi’i werthuso ar lefel uchaf yr integreiddiad llai’r integryn amhendant ar lefel isaf yr integreiddiad.

$$\frac{9^{2}}{2}+\frac{2}{3}\times 9^{\frac{3}{2}}-\left(\frac{4^{2}}{2}+\frac{2}{3}\times 4^{\frac{3}{2}}\right)$$

Symleiddio.

$$\frac{271}{6}$$