Issolvi ∫ (from 0 to 1) of (6t^2+4t+6)(120t+40)dt

$$\int _ { 0 } ^ { 1 } ( 6 t ^ { 2 } + 4 t + 6 ) ( 120 t + 40 ) d t$$

Evalwa

$$1100$$

Passi tas-Soluzzjonijiet

Uża l-propjetà distributtiva biex timmultiplika 6t^{2}+4t+6 b'120t+40 u kkombina termini simili.

$$\int _{0}^{1}720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

L-ewwel evalwa l-integru indefinit.

$$\int 720t^{3}+720t^{2}+880t+240\mathrm{d}t$$

Tintegra s-somma terminu b'terminu.

$$\int 720t^{3}\mathrm{d}t+\int 720t^{2}\mathrm{d}t+\int 880t\mathrm{d}t+\int 240\mathrm{d}t$$

Iffattura ‘l barra l-kostanti f’kull wieħed minn dawn it-termini.

$$720\int t^{3}\mathrm{d}t+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Minn \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int t^{3}\mathrm{d}t ma' \frac{t^{4}}{4}. Immultiplika 720 b'\frac{t^{4}}{4}.

$$180t^{4}+720\int t^{2}\mathrm{d}t+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Minn \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int t^{2}\mathrm{d}t ma' \frac{t^{3}}{3}. Immultiplika 720 b'\frac{t^{3}}{3}.

$$180t^{4}+240t^{3}+880\int t\mathrm{d}t+\int 240\mathrm{d}t$$

Minn \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} għal k\neq -1, issostitwixxi \int t\mathrm{d}t ma' \frac{t^{2}}{2}. Immultiplika 880 b'\frac{t^{2}}{2}.

$$180t^{4}+240t^{3}+440t^{2}+\int 240\mathrm{d}t$$

Sib l-integru ta' 240 billi tuża t-tabella tar-regola tal-integrali komuni \int a\mathrm{d}t=at.

$$180t^{4}+240t^{3}+440t^{2}+240t$$

L-integru definit huwa l-antiderivattiv tal-espressjoni evalwata fil-limitu superjuri tal-integrazzjoni minus l-antiderivattiv evalwat fil-limitu inferjuri tal-integrazzjoni.

$$180\times 1^{4}+240\times 1^{3}+240\times 1+440\times 1^{2}-\left(180\times 0^{4}+240\times 0^{3}+240\times 0+440\times 0^{2}\right)$$

Issimplifika.

$$1100$$