Issolvi ∫ sqrt[3]{3t}dt=(3t)^4/2/4tc

$$\int \sqrt[ 3 ] { 3 t } d t = \frac { ( 3 t ) ^ { 4 / 2 } } { 4 } t c$$

Solvi għal c

$$\left\{\begin{matrix}c=\frac{3^{\frac{4}{3}}}{9t^{\frac{5}{3}}}+\frac{4С}{9t^{3}}\text{, }&t\neq 0\\c\in \mathrm{R}\text{, }&С=0\text{ and }t=0\end{matrix}\right.$$

Passi biex Issolvi l-Ekwazzjoni Lineari

Immultiplika ż-żewġ naħat tal-ekwazzjoni b'4.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{\frac{4}{2}}tc$$

Iddividi 4 b'2 biex tikseb2.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{2}tc$$

Espandi \left(3t\right)^{2}.

$$4\int \sqrt[3]{3t}\mathrm{d}t=3^{2}t^{2}tc$$

Ikkalkula 3 bil-power ta' 2 u tikseb 9.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{2}tc$$

Biex timmultiplika l-qawwa tal-istess bażi, żid l-esponenti tagħhom. Żid 2 u 1 biex tikseb 3.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{3}c$$

Ibdel in-naħat sabiex it-termini varjabbli kollha jkunu fuq in-naħa tax-xellug.

$$9t^{3}c=4\int \sqrt[3]{3t}\mathrm{d}t$$

L-ekwazzjoni hija f'forma standard.

$$9t^{3}c=4\sqrt[3]{3}t^{\frac{4}{3}}+4С$$

Iddividi ż-żewġ naħat b'9t^{3}.

$$\frac{9t^{3}c}{9t^{3}}=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Meta tiddividi b'9t^{3} titneħħa l-multiplikazzjoni b'9t^{3}.

$$c=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Iddividi \frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С b'9t^{3}.

$$c=\frac{4\left(\frac{\left(3t\right)^{\frac{4}{3}}}{3}+С\right)}{9t^{3}}$$