Let lim _{c rightarrow 0} int_c^x frac{b t co | vedclass

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

Question

(B) Let $g(x) = \lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Applying the limit as $x \righ

Vedclass · Accepted Answer

$a = 1 / 4, b = 1$

Vedclass · Answer

(B) Let $g(x) = \lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Applying the limit as $x \rightarrow 0$, we have $g(0) = \lim _{x \rightarrow 0} (\frac{a \sin 4 x}{x} - 1) = 4a - 1$. Since the integral from $c$ to $c$ is $0$, we must have $g(0) = 0$, so $4a - 1 = 0$, which gives $a = 1/4$. Now, differentiate both sides with respect to $x$ using the Fundamental Theorem of Calculus: $g'(x) = \frac{b x \cos 4 x - a \sin 4 x}{x^2}$. Also, differentiating the right side: $g'(x) = \frac{d}{dx} (\frac{a \sin 4 x}{x} - 1) = \frac{4ax \cos 4 x - a \sin 4 x}{x^2}$. Comparing the two expressions for $g'(x)$, we get $b = 4a$. Since $a = 1/4$, we have $b = 4(1/4) = 1$. Thus, $a = 1/4$ and $b = 1$.

Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Find the values of $a$ and $b$.

Similar Questions

If $f(t) = \int_{-t}^t \frac{e^{-|x|}}{2} dx$, then $\lim_{t \rightarrow \infty} f(t)$ is equal to:

If $\theta_{1}$ and $\theta_{2}$ are respectively the smallest and the largest values of $\theta$ in $(0, 2\pi) - \{\pi\}$ which satisfy the equation $2 \cot^{2} \theta - \frac{5}{\sin \theta} + 4 = 0$,then $\int_{\theta_{1}}^{\theta_{2}} \cos^{2} 3\theta \, d\theta$ is equal to

$\int_0^{\pi /2} {\sqrt {\cos \theta } {{\sin }^3}\theta } \,d\theta = $

$\int_0^{\pi / 2} \sin ^5\left(\frac{x}{2}\right) \cdot \sin x \, dx =$

Evaluate $\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module