Datrys f ( x ) = ∫ (from 0 to x) of (t^3+2t^2+1)dt

$$f ( x ) = \int _ { 0 } ^ { x } ( t ^ { 3 } + 2 t ^ { 2 } + 1 ) d t$$

Enrhifo

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

Camau Ateb

Gwerthuso’r integryn amhenodol yn gyntaf.

$$\int t^{3}+2t^{2}+1\mathrm{d}t$$

Integreiddio'r swm fesul term.

$$\int t^{3}\mathrm{d}t+\int 2t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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

$$\int t^{3}\mathrm{d}t+2\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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

$$\frac{t^{4}}{4}+2\int t^{2}\mathrm{d}t+\int 1\mathrm{d}t$$

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

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+\int 1\mathrm{d}t$$

Canfod integryn 1 gan ddefnyddio'r rheol tabl o integrynnau cyffredin \int a\mathrm{d}t=at.

$$\frac{t^{4}}{4}+\frac{2t^{3}}{3}+t$$

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.

$$\frac{x^{4}}{4}+\frac{2}{3}x^{3}+x-\left(\frac{0^{4}}{4}+\frac{2}{3}\times 0^{3}+0\right)$$

Symleiddio.

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

Gwahaniaethu w.r.t. x

$$x^{3}+2x^{2}+1$$