Question

$\int\limits^2_1 {x^{2}-8x+8 } \, dx$

Answer 1

Answer:

${ \underline{ \sf{\int\limits^2_1 {x^{2}-8x+8 } \, dx} \: = \: - 1}}$

Step-by-step explanation:

${ \tt{\int\limits^2_1 {x^{2}-8x+8 } \, dx}} \\ \\ = { \tt{[ \frac{ {x}^{3} }{3} - 4 {x}^{2} + 8x ] {}^{2} _{1}}}$

Substitute x with the limits:

$= { \tt{( \frac{ {2}^{3} }{3} - 4( {2)}^{2} + 8(2)) - ( \frac{ {1}^{3} }{3} - 4( {1)}^{2} + 8(1)) }} \\ = { \tt{( \frac{8}{3} - \frac{11}{3} )}} \\ \\ = { \tt{ - 1}}$