Strange proposition in calculus

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Homework Statement



f(x) is an injective function (1 to 1) and continuous in [a, b], and f(a) < f(b). Show that the
range of f is the interval [f(a), f(b)]

Homework Equations



Intermediate Value Theorem

The Attempt at a Solution


We are asked to use the intermediate value theorem to prove it. However, it seems to me that the proposition is false.

Suppose f(x) = x, a = 0, b = 1. f(x) is 1 to 1, continuous in [a,b] and f(a) < f(b), but its range is (-inf, inf), not [0, 1].

Am I reading this question wrong??
 
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I think the question is asking you to show that the range on [a,b] is [f(a),f(b)]
 
Ah... that makes sense...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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