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

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(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

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$a = 1 / 4, b = 1$

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(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 $b$.

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