Text Compute ∫02(5x3+3x2+2x+1)dx\displaystyle\int_{0}^{2}\left(5x^{3}+3x^{2}+2x+1\right)dx∫02(5x3+3x2+2x+1)dx. Solution An antiderivative is F(x)=54x4+x3+x2+xF(x)=\dfrac{5}{4}x^{4}+x^{3}+x^{2}+xF(x)=45x4+x3+x2+x. By the fundamental theorem: ∫02 (5x3+3x2+2x+1)dx=[54x4+x3+x2+x]02=(20+8+4+2)=34.\int_{0}^{2}\!\left(5x^{3}+3x^{2}+2x+1\right)dx=\left[\tfrac{5}{4}x^{4}+x^{3}+x^{2}+x\right]_{0}^{2}=\left(20+8+4+2\right)=34.∫02(5x3+3x2+2x+1)dx=[45x4+x3+x2+x]02=(20+8+4+2)=34. 34\boxed{34}34