Question

2
Evaluate : ∫ x^3/3x+cos x dx
-2
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Answer 1

Answer:

........................

Answer 2

Answer: $\frac{16}{9}$

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Step-by-step explanation:

$\int\limits^2_ {-2} (\frac{x^{3} }{3x} +cosx )\, dx$

$\int\limits^2_ {-2} (\frac{x^{2} }{3} +cosx )\, dx$

$\int\limits^2_ {-2} \frac{x^{2} }{3} \, dx+\int\limits^2_ {-2} cosx \, dx$

$\frac{x^{3} }{9} +sinx$

$\frac{8}{9}-(-\frac{8}{9})+cos2-cos(-2)$

$\frac{16}{9}+cos2-cos2$

$\frac{16}{9}$