Lutasin ang [3/4*0,001t^4-1/3*0,01t^3-1/20,3t^2+4t]_0^12

$$[ \frac { 3 } { 4 } \cdot 0,001 t ^ { 4 } - \frac { 1 } { 3 } \cdot 0,01 t ^ { 3 } - \frac { 1 } { 2 } 0,3 t ^ { 2 } + 4 t ] _ { 0 } ^ { 12 }$$

I-evaluate

$$-\frac{t^{3}}{300}-\frac{3t^{2}}{20}+4t$$

Mga Hakbang ng Solution

I-multiply ang \frac{3}{4} at 0 para makuha ang 0.

$$0\times 0\times 0\times 1t^{4}-\frac{1}{3}\times 0.01t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

I-multiply ang 0 at 0 para makuha ang 0.

$$0\times 0\times 1t^{4}-\frac{1}{3}\times 0.01t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

I-multiply ang 0 at 0 para makuha ang 0.

$$0\times 1t^{4}-\frac{1}{3}\times 0.01t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

I-multiply ang 0 at 1 para makuha ang 0.

$$0t^{4}-\frac{1}{3}\times 0.01t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

Ang kahit anong imu-multiply sa zero ay zero pa rin.

$$0-\frac{1}{3}\times 0.01t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

I-multiply ang \frac{1}{3} at 0.01 para makuha ang \frac{1}{300}.

$$0-\frac{1}{300}t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

Ang kahit anong idadagdag sa zero ay ganoon pa rin.

$$-\frac{1}{300}t^{3}-\frac{1}{2}\times 0.3t^{2}+4t$$

I-multiply ang \frac{1}{2} at 0.3 para makuha ang \frac{3}{20}.

$$-\frac{1}{300}t^{3}-\frac{3}{20}t^{2}+4t$$

I-factor

$$-\frac{1}{300}t\left(t-\frac{-5\sqrt{273}-45}{2}\right)\left(t-\frac{5\sqrt{273}-45}{2}\right)$$