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Find an equation of the tangent line of the function \( g(x)=4 x+\int_{x}^{\frac{\pi}{4}} \sin (2 t) d t \) at \( x=\frac{\pi}{4} \)

Ask by Chadwick Tyler. in the United Arab Emirates
Mar 17,2025

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The equation of the tangent line at \( x = \frac{\pi}{4} \) is \( y = 3x + \frac{\pi}{4} \).

Solución

To find the equation of the tangent line to the function \[ g(x) = 4x + \int_{x}^{\frac{\pi}{4}} \sin(2t) \, dt \] at \( x = \frac{\pi}{4} \), we need to follow these steps: 1. **Evaluate \( g\left(\frac{\pi}{4}\right) \)** to find the point on the curve. 2. **Find \( g'(x) \)** to determine the slope of the tangent line at that point. 3. **Use the point-slope form of the line** to write the equation of the tangent line. ### Step 1: Evaluate \( g\left(\frac{\pi}{4}\right) \) First, we substitute \( x = \frac{\pi}{4} \) into the function: \[ g\left(\frac{\pi}{4}\right) = 4\left(\frac{\pi}{4}\right) + \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \sin(2t) \, dt \] The integral from \( \frac{\pi}{4} \) to \( \frac{\pi}{4} \) is zero: \[ g\left(\frac{\pi}{4}\right) = \pi + 0 = \pi \] ### Step 2: Find \( g'(x) \) To find the derivative \( g'(x) \), we apply the Fundamental Theorem of Calculus and the derivative of the first term: \[ g'(x) = 4 + \frac{d}{dx} \left( \int_{x}^{\frac{\pi}{4}} \sin(2t) \, dt \right) \] Using the Leibniz rule for differentiation under the integral sign, we have: \[ \frac{d}{dx} \left( \int_{x}^{\frac{\pi}{4}} \sin(2t) \, dt \right) = -\sin(2x) \] Thus, the derivative becomes: \[ g'(x) = 4 - \sin(2x) \] Now, we evaluate \( g'\left(\frac{\pi}{4}\right) \): \[ g'\left(\frac{\pi}{4}\right) = 4 - \sin\left(2 \cdot \frac{\pi}{4}\right) = 4 - \sin\left(\frac{\pi}{2}\right) = 4 - 1 = 3 \] ### Step 3: Write the equation of the tangent line The slope of the tangent line at \( x = \frac{\pi}{4} \) is 3, and the point of tangency is \( \left(\frac{\pi}{4}, \pi\right) \). Using the point-slope form of the line: \[ y - y_1 = m(x - x_1) \] where \( m = 3 \), \( x_1 = \frac{\pi}{4} \), and \( y_1 = \pi \): \[ y - \pi = 3\left(x - \frac{\pi}{4}\right) \] Simplifying this gives: \[ y - \pi = 3x - \frac{3\pi}{4} \] \[ y = 3x - \frac{3\pi}{4} + \pi \] \[ y = 3x + \frac{\pi}{4} \] Thus, the equation of the tangent line is: \[ \boxed{y = 3x + \frac{\pi}{4}} \]

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To find the equation of the tangent line for the function \( g(x) = 4x + \int_{x}^{\frac{\pi}{4}} \sin(2t) dt \) at \( x = \frac{\pi}{4} \), we need to first compute \( g\left(\frac{\pi}{4}\right) \) and \( g'\left(x\right) \). 1. First, we evaluate \( g\left(\frac{\pi}{4}\right) \): \[ g\left(\frac{\pi}{4}\right) = 4\left(\frac{\pi}{4}\right) + \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \sin(2t) dt = \pi + 0 = \pi. \] 2. Next, we find the derivative \( g'(x) \): By the Fundamental Theorem of Calculus, we have: \[ g'(x) = 4 + \frac{d}{dx} \int_{x}^{\frac{\pi}{4}} \sin(2t) dt = 4 - \sin(2x). \] Now, evaluate \( g'\left(\frac{\pi}{4}\right) \): \[ g'\left(\frac{\pi}{4}\right) = 4 - \sin\left(2 \times \frac{\pi}{4}\right) = 4 - \sin\left(\frac{\pi}{2}\right) = 4 - 1 = 3. \] 3. Now, we have a point \(\left(\frac{\pi}{4}, \pi\right)\) and a slope of 3. The equation of the tangent line in point-slope form is: \[ y - \pi = 3\left(x - \frac{\pi}{4}\right). \] Simplifying this gives: \[ y = 3x - \frac{3\pi}{4} + \pi = 3x + \frac{\pi}{4}. \] Thus, the equation of the tangent line is: \[ y = 3x + \frac{\pi}{4}. \]

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