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Integral Calculus

Blazing goIITian

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29 Mar 2009 18:58:22 IST

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Integrate

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akhil

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Joined: 26 Feb 2009

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29 Mar 2009 19:00:33 IST

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question ?

Srujana

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30 Mar 2009 09:46:33 IST

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$\int_{1}^{2}{x^{[x^2]}}\ dx+\int_{1}^{3}[x^2]^x\ dx$

$Let\ x^2=t\\ \\ \Rightarrow dx=\frac{dt}{2}$

$\frac{1}{2}\int_{1}^{4}{\sqrt{t}^{\ [t]}}\ dt+\frac{1}{2}\int_{1}^{9}[t]^{\sqrt{t}}\ dx$

$=\frac{1}{2}\left({\int_{1}^{2}\sqrt{t}\ dt}+\int_{2}^{3}t\ dt+\int_{3}^{4}\sqrt{t}^{\ 3}dt\right)$ $+\frac{1}{2}\left(\int_{1}^{2}[t]^{\sqrt{t}}dt+\int_{2}^{3}[t]^{\sqrt{t}}dt+ ...+\int_{8}^{9}[t]^{\sqrt{t}}dt \right)$
$=\frac{1}{2}\left({\int_{1}^{2}\sqrt{t}\ dt}+\int_{2}^{3}t\ dt+\int_{3}^{4}\sqrt{t}^{\ 3}dt\right)$ $+\frac{1}{2}\left(1+\int_{2}^{3}2^{\sqrt{t}}dt+\int_{3}^{4}3^{\sqrt{t}}dt+...+\int_{8}^{9}8^{\sqrt{t}}dt\right)$

It can be easily integrated from this point

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