Q: If α ε (2,3) , Find the no of solutions of \int_{0}^{\alpha }{cos(x+\alpha ^{2})dx} = sin\alpha
Q: Let f : [0,1] → R is a continuous function such that \int_{0}^{1}{f(x) dx=0}.
Prove that there is some ' c ' ε (0,1) such that \int_{0}^{c}{f(x) dx} = f(c)
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10 Answers
Q: \int_{0}^{\frac{\pi }{4} }{(tan^{n}x+tan^{n-2}x)}d(x-[x])
where, [.] is step function.
Attempting the first one:
\text{Let } g(x)=\int_{0}^{x}{f(x)} dx
\Rightarrow g'(x)=f(x)
\text{Given } \int_{0}^{1}{f(x)}dx=0 \Rightarrow g(0)=g(1)
Since the function is continuous applying Rolle's Theorem in (0,1),
g'(c)=0 \text{ for some } c \in (0,1)
I'm seem to be stuck here. Don't know if this leads anywhere!
i think the in first q. the limits of the integral should be 0→1. instead of 0→c
1) Let for each x in [0,1],
F(x) = \int_0^x f(t)\ \mathrm dt
and further define a function
g(x) = e^{-x}F(x)\quad x\in[0,1]
Since f(x) is continuous in [0,1], so F(x) is differentiable and hence g(x) is differentiable in (0,1). At the same time g(0) = g(1) =0. Hence according to Rolle's theorem there exist at least one c in (0,1) so that g'(c) =0.
But
g'(x) = e-x F'(x) - e-x F(x) = e-x (F'(x) - F(x))
Since e-x ≠0, therefore there is atleast one c in (0,1) such that
F'(c) = F(c)
i.e.
f(c) = \int_0^c f(t)\mathrm dt
For 2). The given equation is equivalent to
\int_0^\alpha (\cos (x+\alpha^2) - \cos x)\mathrm dx =0
i.e.
-2\sin(\alpha^2/2)\int_0^\alpha \sin\left(x+\frac{\alpha^2}{2}\right)\mathrm dx = 0
This gives either
\sin \left(\frac{\alpha^2}{2}\right) = 0 ---- \quad (1)
or
\int_0^\alpha \sin\left(x+\frac{\alpha^2}{2}\right)\mathrm dx = 0 ---\quad (2)
From (1), we get
\alpha^2 = 2n \pi for some integer n.
Since \alpha lies between 2 and 3, its square will lie between 4 and 9 and so the only possibility is that \alpha^2 = 2 \pi giving
\boxed{\alpha = \sqrt{2\pi}}
From (2) we get
\cos\left(\alpha + \frac{\alpha^2}{2}\right) = \cos \left(\frac{\alpha^2}{2}\right)
So that
\alpha + \frac{\alpha^2}{2} = 2k\pi \pm \frac{\alpha^2}{2}
for integer k.
Taking the plus sign, we get \alpha = 2k\pi which cannot be true for any integral k (since alpha lies between 2 and 3).
Taking the minus sign, we get
\alpha + \alpha^2 = 2k\pi
Since 6<\alpha + \alpha^2 < 12, the only possibility is that
\alpha + \alpha^2 = 2\pi i.e. \alpha^2 + \alpha -2\pi = 0 which has got one solution in (2,3) which is
\alpha = \frac{1}{2}\left(\sqrt{1+8\pi}-1\right)
And hence the original equation has exactly two roots in (2,3).
3) For the given interval x-[x] = x and hence the given integral is simply
\int_0^{\pi/4} (\tan^nx + \tan^{n-2}x)\mathrm dx
which is quite trivially equal to
\frac{1}{n-1}
for n≠1.