Evaluate [tex] \int \limits^5_2 { x^{2} + 3x } \, dx [/tex].
Answer (2)
2 ∫ 5 x 2 + 3 x d x = [ 3 x 3 + 2 3 x 2 ] 2 5 = 3 5 3 + 2 3 ⋅ 5 2 − ( 3 2 3 + 2 3 ⋅ 2 2 ) = 3 125 + 2 75 − ( 3 8 + 6 ) = 6 250 + 6 225 − ( 6 16 + 6 36 ) = 6 475 − 6 52 = 6 423 = 2 141 = 70.5
To evaluate the integral ∫ 2 5 ( x 2 + 3 x ) d x , we found the antiderivative, calculated it at the upper and lower limits, and determined the final result to be 6 423 or approximately 70.5 .
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