Let S(x) = int_{x^2}^{x^3} ln t , dt for x | vedclass

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

Question

(B) Using the Leibniz integral rule,we differentiate $S(x)$ with respect to $x$: $S'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{d

Vedclass · Accepted Answer

derivable and continuous in its domain

Vedclass · Answer

(B) Using the Leibniz integral rule,we differentiate $S(x)$ with respect to $x$: $S'(x) = \ln(x^3) \cdot \frac{d}{dx}(x^3) - \ln(x^2) \cdot \frac{d}{dx}(x^2)$ $S'(x) = (3 \ln x) \cdot (3x^2) - (2 \ln x) \cdot (2x)$ $S'(x) = 9x^2 \ln x - 4x \ln x$ $S'(x) = x \ln x (9x - 4)$ Now,$H(x) = \frac{S'(x)}{x} = \frac{x \ln x (9x - 4)}{x} = \ln x (9x - 4)$ Since $\ln x$ and $(9x - 4)$ are both continuous and differentiable for all $x > 0$,their product $H(x) = \ln x (9x - 4)$ is also continuous and differentiable in its domain $(0, \infty)$.

Let $S(x) = \int_{x^2}^{x^3} \ln t \, dt$ for $x > 0$ and $H(x) = \frac{S'(x)}{x}$. Then $H(x)$ is :

Similar Questions

The points of extremum of $\int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$ are

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ is equal to

$\int_0^\pi x \cdot \sin^5 x \cdot \cos^6 x \, dx =$

If $\int_{\sin x}^1 {{t^2}f(t)\;dt = 1 - \sin x} $,$x \in \left( {0,\frac{\pi }{2}} \right)$,then $f\left( {\frac{1}{{\sqrt 3 }}} \right)$ is equal to:

$\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module