Réitigh ∫ (from 1 to 5) of (15t^3-135t^2+225t)dt=

$$\int _ { 1 } ^ { 5 } ( 15 t ^ { 3 } - 135 t ^ { 2 } + 225 t ) d t =$$

Luacháil

$$-540$$

Céimeanna Réitigh

Déan luacháil ar an suimeálaí éiginnte ar dtús.

$$\int 15t^{3}-135t^{2}+225t\mathrm{d}t$$

Measc an tsuim téarma fá téarma.

$$\int 15t^{3}\mathrm{d}t+\int -135t^{2}\mathrm{d}t+\int 225t\mathrm{d}t$$

Fág an leanúnach sna téarmaí as an áireamh.

$$15\int t^{3}\mathrm{d}t-135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t$$

Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int t^{3}\mathrm{d}t le \frac{t^{4}}{4}. Méadaigh 15 faoi \frac{t^{4}}{4}.

$$\frac{15t^{4}}{4}-135\int t^{2}\mathrm{d}t+225\int t\mathrm{d}t$$

Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int t^{2}\mathrm{d}t le \frac{t^{3}}{3}. Méadaigh -135 faoi \frac{t^{3}}{3}.

$$\frac{15t^{4}}{4}-45t^{3}+225\int t\mathrm{d}t$$

Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int t\mathrm{d}t le \frac{t^{2}}{2}. Méadaigh 225 faoi \frac{t^{2}}{2}.

$$\frac{15t^{4}}{4}-45t^{3}+\frac{225t^{2}}{2}$$

Is ionann suimeálaí cinnte agus frithdhíorthach an nath luacháilte ag teorainn uachtair na suimeála lúide an frithdhíorthach luacháilte ag teorainn íochtair na suimeála.

$$\frac{15}{4}\times 5^{4}-45\times 5^{3}+\frac{225}{2}\times 5^{2}-\left(\frac{15}{4}\times 1^{4}-45\times 1^{3}+\frac{225}{2}\times 1^{2}\right)$$

Simpligh.

$$-540$$