Evaluate, \( \int_{1}^{5}\left(\frac{3 x}{x^{3}}\right) d x \)
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To evaluate the integral \( \int_{1}^{5}\left(\frac{3 x}{x^{3}}\right) d x \), we first simplify the integrand: \[ \frac{3x}{x^3} = \frac{3}{x^2} \] So, we can rewrite the integral as: \[ \int_{1}^{5} \frac{3}{x^2} \, dx \] Next, we recall that the integral of \( x^{-2} \) is: \[ \int x^{-2} \, dx = -\frac{1}{x} + C \] Applying this to our integral, we have: \[ \int \frac{3}{x^2} \, dx = 3 \left(-\frac{1}{x}\right) = -\frac{3}{x} + C \] Now, we evaluate the definite integral from 1 to 5: \[ \left[-\frac{3}{x}\right]_{1}^{5} = \left(-\frac{3}{5}\right) - \left(-\frac{3}{1}\right) = -\frac{3}{5} + 3 \] Converting 3 to a fraction with a denominator of 5: \[ 3 = \frac{15}{5} \] So the calculation now looks like this: \[ -\frac{3}{5} + \frac{15}{5} = \frac{12}{5} \] Thus, the value of the integral is: \[ \int_{1}^{5}\left(\frac{3 x}{x^{3}}\right) d x = \frac{12}{5} \]
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