Solve [0001t^3-001t^2-03t+4]_0^12

$$[ 0001 t ^ { 3 } - 001 t ^ { 2 } - 03 t + 4 ] _ { 0 } ^ { 12 }$$

Evaluate

$$\text{Indeterminate}$$

Solution Steps

Multiply 0 and 0 to get 0.

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

Multiply 0 and 0 to get 0.

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

Multiply 0 and 1 to get 0.

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

Anything times zero gives zero.

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

Multiply 0 and 0 to get 0.

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

Multiply 0 and 1 to get 0.

$$0-0t^{2}-0\times 3t+4$$

Anything times zero gives zero.

$$0-0-0\times 3t+4$$

Subtracting 0 from itself leaves 0.

$$0-0\times 3t+4$$

Multiply 0 and 3 to get 0.

$$0-0t+4$$

Anything times zero gives zero.

$$0-0+4$$

Subtracting 0 from itself leaves 0.

$$0+4$$

Add 0 and 4 to get 4.

$$4$$