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Evaluate $\int_1^2 (x + 4)^2 dx $

$$\eqalign{ \int_1^2 (x + 4)^2 dx = \int_1^2 (x^2 + 8x + 16) dx \ &= \left\lbrack {x^3 \over 3} + {8x^2 \over 2} + 16x \right\rbrack_1^2 \ &= \left\lbrack {8 \over 3} + {8 * 4 \over 2} + 16 * 2 \right\rbrack - \left\lbrack {1 \over 3} + {8 \over 2} + 16 \right\rbrack }$$