Let \(I=\int \sqrt{x^2+4x-5}dx\)
=\(\int\sqrt{(x^2+4x+4)-9}dx\)
\(=\int \sqrt{(x+2)^2-(3)^2}dx\)
It is known that,\(\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\mid x+\sqrt{x^2-a^2}\mid+C\)
∴\(I= \frac{(x+2)}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\mid (x+2)+\sqrt{x^2+4x-5}\mid +C\)
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)
मोबाइल फोन विहीन दुनिया — 120 शब्दों में रचनात्मक लेख लिखिए :
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
These are tabulated below along with the meaning of each part.
