int_1^4 left(x + sqrt{x} + frac{1}{x}right) d | vedclass

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

Question

(A) Let $I = \int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx$. First, evaluate the first integral: $\int_1^4 \left(x + x^{1

Vedclass · Accepted Answer

$\frac{79}{6}$

Vedclass · Answer

(A) Let $I = \int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx$. First, evaluate the first integral: $\int_1^4 \left(x + x^{1/2} + \frac{1}{x}\right) dx = \left[ \frac{x^2}{2} + \frac{2}{3}x^{3/2} + \log |x| \right]_1^4$. $= \left( \frac{16}{2} + \frac{2}{3}(8) + \log 4 \right) - \left( \frac{1}{2} + \frac{2}{3} + 0 \right) = \left( 8 + \frac{16}{3} + 2 \log 2 \right) - \left( \frac{7}{6} \right) = \frac{40}{3} + 2 \log 2 - \frac{7}{6} = \frac{80 - 7}{6} + 2 \log 2 = \frac{73}{6} + 2 \log 2$. Now, evaluate the second integral: $\int_1^{2 \log 2} dx = [x]_1^{2 \log 2} = 2 \log 2 - 1$. Subtracting the two results: $I = \left( \frac{73}{6} + 2 \log 2 \right) - (2 \log 2 - 1) = \frac{73}{6} + 1 = \frac{73 + 6}{6} = \frac{79}{6}$.

$\int_1^4 \left(x + \sqrt{x} + \frac{1}{x}\right) dx - \int_1^{2 \log 2} dx = $

Similar Questions

The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$,where $[.]$ denotes the greatest integer function,is equal to:

The approximate value of $\int_{1}^{9} x^2 dx$ by using the trapezoidal rule with $4$ equal intervals is:

$\int_0^3 |x^2 - 3x + 2| dx = $

For $0 < x < \frac{\pi}{2}$,the integral $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$ is equal to:

$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module