find-0-2-x-2-x-2-dx-

Question Number 90171 by M±th+et£s last updated on 21/Apr/20

f i n d \int_{0}^{2} (⌊ x^{2} ⌋ + ⌊ x ⌋^{2}) d x

Answered by TANMAY PANACEA. last updated on 21/Apr/20

i s i t f l o o r f u n c t i o n (g r e a t e s t i n t e g e r f u n c t i o n 3

Commented by M±th+et£s last updated on 21/Apr/20

y e s s i r

Answered by TANMAY PANACEA. last updated on 21/Apr/20

\int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{\sqrt{3}} [x^{2}] d x + \int_{\sqrt{3}}^{2} [x^{2}] d x

= \int_{0}^{1} 0 . d x + \int_{1}^{\sqrt{2}} 1 . d x + \int_{\sqrt{2}}^{\sqrt{3}} 2 . d x + \int_{\sqrt{3}}^{2} 3 . d x

= 0.2) + 1 . (\sqrt{2} - 1) + 2 (\sqrt{3} - \sqrt{2}) + 3 (2 - \sqrt{3})

= 0 + \sqrt{2} - 1 + 2 \sqrt{3} - 2 \sqrt{2} + 6 - 3 \sqrt{3}

= 5 - \sqrt{2} - \sqrt{3}

\int_{0}^{2} {[x]}^{2} d x

\int_{0}^{1} 0 \times d x + \int_{1}^{2} 1 \times d x

= 1

A n s w e r = 5 - \sqrt{2} - \sqrt{3} + 1

= 6 - \sqrt{2} - \sqrt{3}

Commented by M±th+et£s last updated on 21/Apr/20

n i c e w o r k s i r t h a n k y o u

Commented by TANMAY PANACEA. last updated on 21/Apr/20

m o s t w e l c o m e s i r