intlimits_0^3 left( frac{1}{sqrt{x^2 + 4x + 4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The integral is $I = \int_0^3 \left( \frac{1}{\sqrt{(x+2)^2}} + \sqrt{(x-2)^2} \right) dx = \int_0^3 \left( \frac{1}{|x+2|} + |x-2| \right) dx$. S

Vedclass · Accepted Answer

$\ln \frac{5}{2} + \frac{5}{2}$

Vedclass · Answer

(C) The integral is $I = \int_0^3 \left( \frac{1}{\sqrt{(x+2)^2}} + \sqrt{(x-2)^2} \right) dx = \int_0^3 \left( \frac{1}{|x+2|} + |x-2| \right) dx$. Since $x \in [0, 3]$,$|x+2| = x+2$. Thus,$I = \int_0^3 \frac{1}{x+2} dx + \int_0^3 |x-2| dx$. Evaluating the first part: $\int_0^3 \frac{1}{x+2} dx = [\ln|x+2|]_0^3 = \ln 5 - \ln 2 = \ln \frac{5}{2}$. Evaluating the second part: $\int_0^3 |x-2| dx = \int_0^2 -(x-2) dx + \int_2^3 (x-2) dx$. $= [-\frac{x^2}{2} + 2x]_0^2 + [\frac{x^2}{2} - 2x]_2^3$. $= (-2 + 4) + ((\frac{9}{2} - 6) - (2 - 4)) = 2 + (-\frac{3}{2} + 2) = 2 + \frac{1}{2} = \frac{5}{2}$. Therefore,$I = \ln \frac{5}{2} + \frac{5}{2}$.

$\int\limits_0^3 \left( \frac{1}{\sqrt{x^2 + 4x + 4}} + \sqrt{x^2 - 4x + 4} \right) dx =$

Similar Questions

$\int_{\pi / 6}^{\pi / 3} \cos^{-4} x \, dx =$

$\int_2^3 \frac{d x}{x^2-x}$ is equal to

Let $\alpha \in (0,1)$ and $\beta = \log_{e}(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^n}{n}$ for $x \in (0,1)$. Then the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} dt$ is equal to

Domain of definition of the function $f(x) = \int\limits_0^x \frac{dt}{\sqrt{x^2 + t^2}}$ is

If $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$,where $n$ is a positive integer,then $I_{10}+I_{8}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module