Answer
$$\int^1_{-1}(3x^2-4x+7)dx=16$$
Work Step by Step
$$A=\int^1_{-1}(3x^2-4x+7)dx$$ $$A=\Big(\frac{3x^3}{3}-\frac{4x^2}{2}+7x\Big)\Big]^1_{-1}$$ $$A=\Big(x^3-2x^2+7x\Big)\Big]^1_{-1}$$ $$A=(1^3-2\times1^2+7\times1)-\Big((-1)^3-2\times(-1)^2+7\times(-1)\Big)$$ $$A=(1-2+7)-(-1-2-7)$$ $$A=6+10=16$$
