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Chapter 5: Problem 34

Evaluate the following integrals using the Fundamental Theorem of Calculus. $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t$$

Short Answer

Expert verified
Question: Evaluate the integral $\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t$. Answer: The value of the integral is $\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t = 2\ln\left(\frac{9}{4}\right) + 2$.

Step by step solution

01

Define the integral

We are given the integral: $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t$$ This is a definite integral because it has limits 4 and 9, and we must find the area under the curve of the function between these limits.
02

Find the antiderivative

To find the antiderivative of the function, we first break it into two simpler fractions: $$\int \frac{2+\sqrt{t}}{t} d t = \int \frac{2}{t} d t + \int\frac{\sqrt{t}}{t} d t$$ Now, we can rewrite the second fraction as \(\int\frac{t^{1/2}}{t} dt = \int{t^{-1/2}} dt\) Now, we can find the antiderivative of each fraction: For the first fraction: \(\int \frac{2}{t} dt = 2 \int \frac{1}{t} dt = 2 \ln |t| + C_1\) For the second fraction: \(\int t^{-1/2} dt = 2t^{1/2} + C_2\) The antiderivative (indefinite integral) of the function is therefore: $$F(t) = 2 \ln |t| + 2t^{1/2} + C$$
03

Apply the Fundamental Theorem of Calculus

We now use the Fundamental Theorem of Calculus by plugging in the upper and lower limits and taking their difference: $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t = F(9) - F(4) = \left(2\ln |9| + 2\cdot9^{1/2}\right) - \left(2\ln |4| + 2\cdot4^{1/2}\right)$$ Now, we calculate numeric values: $$\left(2\ln 9 + 2\cdot3\right) - \left(2\ln 4 + 2\cdot2\right) = (2\ln 9 + 6) - (2\ln 4 + 4)$$ Simplifying gives: $$2(\ln 9 - \ln 4) + 2 = 2\ln\left(\frac{9}{4}\right) + 2$$ So, the value of the integral is: $$\int_{4}^{9} \frac{2+\sqrt{t}}{t} d t = 2\ln\left(\frac{9}{4}\right) + 2$$

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