Datrys ∫ (from 1 to 3) of (x^4+7x^3+2x^2+x+3) wrt x

$$\int _ { 1 } ^ { 3 } ( x ^ { 4 } + 7 x ^ { 3 } + 2 x ^ { 2 } + x + 3 ) d x$$

Enrhifo

$$\frac{3236}{15}\approx 215.733333333$$

Camau Ateb

Gwerthuso’r integryn amhenodol yn gyntaf.

$$\int x^{4}+7x^{3}+2x^{2}+x+3\mathrm{d}x$$

Integreiddio'r swm fesul term.

$$\int x^{4}\mathrm{d}x+\int 7x^{3}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Ffactoreiddio allan y cysonyn ym mhob un o'r termau.

$$\int x^{4}\mathrm{d}x+7\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{4}\mathrm{d}x gyda \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}+7\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

Ers \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} ar gyfer k\neq -1, disodli \int x^{3}\mathrm{d}x gyda \frac{x^{4}}{4}. Lluoswch 7 â \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+2\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\int x\mathrm{d}x+\int 3\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\frac{x^{2}}{2}+\int 3\mathrm{d}x$$

Canfod integryn 3 gan ddefnyddio'r rheol tabl o integrynnau cyffredin \int a\mathrm{d}x=ax.

$$\frac{x^{5}}{5}+\frac{7x^{4}}{4}+\frac{2x^{3}}{3}+\frac{x^{2}}{2}+3x$$

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.

$$\frac{3^{5}}{5}+\frac{7}{4}\times 3^{4}+\frac{2}{3}\times 3^{3}+\frac{3^{2}}{2}+3\times 3-\left(\frac{1^{5}}{5}+\frac{7}{4}\times 1^{4}+\frac{2}{3}\times 1^{3}+\frac{1^{2}}{2}+3\times 1\right)$$

Symleiddio.

$$\frac{3236}{15}$$