Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [\(\frac{1}{n+1},\frac{1}{n}\)] where n ∈ N. Let g : (0,1) → R be a function such that \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt x\) for all x ∈ (0,1).Then \(\lim_{x\rightarrow0}f(x)g(x)\)
∫ (x2 - 1)dx/(x3(2x4 - 2x2 1)1/2) = ?
\(\int \frac{1}{(x + 2)(1 + x)^2} \, dx\)
\(\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} \, dx\)
\( \int \left( \frac{1}{7} - 6x - x^2 \right) \, dx \)
Let n ≥ 2 be a natural number and f:[0,1)]\(\rightarrow\) R be the function defined by\(f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\ 2n((2nx-1)\ \ \ \ \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \end{cases}\)If n is such that the area of the region bounded by the curves x = 0, x = 1, y = 0 and y = f(x) is 4, then the maximum value of the function f is
Calculate the value of ∫1 (tan-1x/1+x2) dx
-{tan(1/x) - (1/x)} + c
∫ex (1 - cotx + cot2x) dx = ?
∫ (dx/(sinx + cosx)).dx = ?