\(I = ∫ \frac{(x^2+1)e^x}{(x+1)^2}dx = ƒ(x)e^x+C\)
\(I = ∫ \frac{e^x(x^2-1+1+1)}{(x+1)^2}dx\)
= \(∫ e^x\bigg[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\bigg]dx\)
= \(e^x\bigg(\frac{x-1}{x+1}\bigg)+c\)
\(∴ f (x) = \frac{x-1}{x+1}\)
\(f(x) = \frac{1- 2}{x+1}\)
\(f' (x) = 2\bigg(\frac{1}{x+1}\bigg)^2\)
\(f''(x) = -4\bigg(\frac{1}{x+1}\bigg)^3\)
\(f'''(x) = \frac{12}{(x+1)^4}\)
for \(x = 1\)
\(f'''(1) = \frac{12}{24}\)
= \(\frac{12}{16}\)
= \(\frac{3}{4}\)
Hence, the correct option is (B): \(\frac{3}{4}\)
The area of the region given by \(\left\{(x, y): x y \leq 8,1 \leq y \leq x^2\right\}\) is :
If 5f(x) + 4f (\(\frac{1}{x}\)) = \(\frac{1}{x}\)+ 3, then \(18\int_{1}^{2}\) f(x)dx is:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
There are distinct applications of integrals, out of which some are as follows:
In Maths
Integrals are used to find:
In Physics
Integrals are used to find: