Question

Evaluate the definite integrals listed below

a. ∫ (e,6) (ln(x))/x dx

b. ∫ (-1,0) x/(x + 2) dx

c. ∫ (0,3) xe^(2x^2) dx

with the first bracketed number being on the bottom of the integral symbol and the second being on the top

Answer 1

Answer on Question #48165 – Math – Integral Calculus

Question: evaluate the definite integrals listed below

$$\int_{e}^{6} \frac{\ln(x)}{x} \, dx$$ $$\int_{-1}^{0} \frac{x}{x + 2} \, dx$$ $$\int_{0}^{3} x \cdot e^{2x^2} \, dx$$

Solution:

1) Let us change the variable of integration:

$$\int_{e}^{6} \frac{\ln(x)}{x} \, dx = \int_{e}^{6} \ln(x) \, d(\ln(x)) = \left. \frac{\ln^2(x)}{2} \right|_{e}^{6} = \frac{\ln^2(6)}{2} - \frac{\ln^2(e)}{2} = \frac{\ln^2(6)}{2} - \frac{1}{2}$$

2) In this case let us use the following trick:

$$\begin{array}{l} \int_{-1}^{0} \frac{x}{x + 2} \, dx = \int_{-1}^{0} \frac{x + 2 - 2}{x + 2} \, dx = \int_{-1}^{0} \left(1 - \frac{2}{x + 2}\right) dx = (x - 2 \ln(x + 2)) \big|_{-1}^{0} \\ = -2 \ln(2) - (-1 - 2 \ln(1)) = 1 - 2 \ln(2) \end{array}$$

3) Here we are also going to change the variable of integration:

$$\int_{0}^{3} x \cdot e^{2x^2} \, dx = \frac{1}{2} \int_{0}^{3} e^{2x^2} \, dx^2 = \frac{1}{4} \int_{0}^{3} e^{2x^2} \, d(2x^2) = \left. \frac{1}{4} e^{2x^2} \right|_{0}^{3} = \frac{1}{4} (e^{18} - 1)$$

Answer:

$$\int_{e}^{6} \frac{\ln(x)}{x} \, dx = \frac{\ln^2(6)}{2} - \frac{1}{2}$$ $$\int_{-1}^{0} \frac{x}{x + 2} \, dx = 1 - 2 \ln(2)$$ $$\int_{0}^{3} x \cdot e^{2x^2} \, dx = \frac{1}{4} (e^{18} - 1)$$

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