3
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Let $f: (0,\infty) \to \mathbb{R}$ be given by $$f(x)=\int_{1/x}^x e^{t+1/t} \frac{dt}{t}$$ then $$\int_{-1}^1 f(2^x)dx$$ is

(a) 1

(b) 2

I tried to solve this problem by differentiated but I didn't proceed so please help how shall I think about this help Thanks .

  • 0
    Show your solution – 2017-01-17
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    Okk I will edit my question then pls help me how I shall I proceed.. – 2017-01-17
  • 0
    Did I reproduce your problem correctly in my edit? – 2017-01-22

1 Answers 1

3

In general, we have $$\int_a^b f(x) \mathrm{d}x = \int_a^b f(a+b-x) \mathrm{d}x$$.

Therefore, in this case, we should have: $$\int_{-1}^1 f(2^x) \mathrm{d}x = \int_{-1}^1 f(2^{-x}) \mathrm{d}x$$

Hence,

$$\int_{-1}^1 f(2^x) \mathrm{d}x = \frac{1}{2}\int_{-1}^1 f(2^x) + f(2^{-x}) \mathrm{d}x $$

But clearly,

\begin{align*} f(2^x) + f(2^{-x}) &= \int_{2^{-x}}^{2^x} e^{t+\frac{1}{t}}\frac{1}{t} \mathrm{d}t + \int_{2^{x}}^{2^{-x}} e^{t+\frac{1}{t}}\frac{1}{t} \mathrm{d}t \\ &= \int_{2^{-x}}^{2^x} e^{t+\frac{1}{t}}\frac{1}{t} \mathrm{d}t - \int_{2^{-x}}^{2^{x}} e^{t+\frac{1}{t}}\frac{1}{t} \mathrm{d}t \\ &= 0 \end{align*}

Therefore, $$\int_{-1}^1 f(2^x) \mathrm{d}x = 0$$

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    Pleasure was all mine! – 2017-01-17