The integral $$I=$$∫−$$3/23/2$$ [x]$$+x3+loga$$⁡ $$x+x2+1dx$$ is equal

Question

The integral $$I=$$∫−$$3/23/2$$ [x]$$+x3+loga$$⁡ $$x+x2+1dx$$ is equal

Infinity Learn · Accepted Answer

Let                            $$f(x)=x3+loga$$⁡ $$x+x2+1$$                             $$f($$−$$x)=$$−$$x3+loga$$⁡ −$$x+x2+1=$$−$$x3+loga$$⁡ $$1x+x2+1=$$−$$x3+loga$$⁡ $$x+x2+1=$$−$$f(x)$$∴ ∫−$$3/23/2$$ $$f(x)dx=0Thus$$                     $$I=$$∫−$$3/23/2$$ [x]dx                             $$=$$∫−$$3/2$$−$$1$$ $$($$−$$2)dx+$$∫−$$10$$ −$$1dx+$$∫$$01$$ $$0dx+$$∫$$13/2$$ $$1dx=($$−$$2)$$[−$$1+3/2$$]$$+($$−$$1)$$[$$0+1$$]$$+0+1$$[$$3/2$$−$$1$$]$$=$$−$$1$$−$$1+1/2=$$−$$3/2$$.

The integral $I = \int_{- 3 / 2}^{3 / 2} ([x] + x^{3} + \log_{a} (x + \sqrt{x^{2} + 1})) d x$ is equal

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The integral I=∫−3/23/2 [x]+x3+loga⁡ x+x2+1dx is equal

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The integral $I = \int_{- 3 / 2}^{3 / 2} ([x] + x^{3} + \log_{a} (x + \sqrt{x^{2} + 1})) d x$ is equal