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If \( \int_{1}^{2} e^{x^{2}} d x=a \) then \( \int_{e}^{e^{4}} \sqrt{\log x} d x \) is equal. |
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Answer» \(\int\limits_e^{e^4} \sqrt{log\, x}\,dx\) Let \(log \,x = t^2 \) ⇒ \(x = e^{t^2}\) \(\frac 1x dx = 2t\,dt\) \(dx = 2t.e^{t^2}dt\) Then \(\int\limits_e^{e^4} \sqrt{log\, x}\,dx = \int\limits_1^2 t.(2te^{t^2})dt\) \(=2\left[\int\limits_1^2 t^2.e^{t^2}dt\right]\) \(= 2\left[[t^2]^2_1 \int\limits_1^2 e^{t^2} dt -\int\limits_1^2 (2t\int e^{t^2} dt)\right]\) \(=2\left[(4-1).a - [2t]_1^2 \int\limits_1^2 a\,dt + \int\limits_1^22.a\right]\) \(\left(\because \int\limits_1^2 et^2dt = a\right)\) \(= 2\left[3a - (4 -1)a (2 -1) + 2a(2 -1)\right]\) \(= 2(3a - 3a + 2a)\) \(= 4a\) |
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