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Evaluate `int^6_3(2x^2+3x+5)dx`. |
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Answer» `int(2x^2+3x+5)dx` ` =int 2x^2dx+int 3xx+int 5 dx` `=2int x^2dx+3intx dx+5 int x^0dx` ` 2x^3/3+3x^2/2+tx^1/1` ` =2/3 x^3+3/2 x^2+5x` Thus, `int^6_3(2x^2+3x+5)dx=[2/3x^3+3/2x^2+5x]^6_3` `2/3(216-27)+3/2(36-9)+5(6-3)` `=126+40.5+15=18.15`. |
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