The value of int_{-1}^1 frac{(1+sqrt{|x|-x}) | vedclass

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Question

(D) Let $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx

Vedclass · Accepted Answer

$1+\frac{2 \sqrt{2}}{3}$

Vedclass · Answer

(D) Let $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} dx$.Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we replace $x$ with $-x$:$I = \int_{-1}^1 \frac{(1+\sqrt{|-x|-(-x)}) e^{-x}+(\sqrt{|-x|-(-x)}) e^{-(-x)}}{e^{-x}+e^{-(-x)}} dx$$I = \int_{-1}^1 \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^x}{e^{-x}+e^x} dx$.Adding the two expressions for $I$:$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x + (\sqrt{|x|-x}) e^{-x} + (1+\sqrt{|x|+x}) e^{-x} + (\sqrt{|x|+x}) e^x}{e^x+e^{-x}} dx$$2I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}+\sqrt{|x|+x})(e^x+e^{-x})}{e^x+e^{-x}} dx$$2I = \int_{-1}^1 (1+\sqrt{|x|-x}+\sqrt{|x|+x}) dx$.Since the integrand is an even function,$2I = 2 \int_0^1 (1+\sqrt{x-x}+\sqrt{x+x}) dx = 2 \int_0^1 (1+0+\sqrt{2x}) dx$.$I = \int_0^1 (1+\sqrt{2}x^{1/2}) dx = [x + \sqrt{2} \cdot \frac{2}{3} x^{3/2}]_0^1$$I = 1 + \frac{2\sqrt{2}}{3}$.

The value of $\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x$ is equal to

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