Use the Fundamental Theorem of Calculus, Part 2, to evaluate the definite integral: int1^e 1/x\,dx What is the exact value of this integral?

Since d/dx[ln x] = 1/x, an antiderivative of f(x) = 1/x is F(x) = ln x. By FTC Part 2, int1^e 1/x\,dx = F(e) - F(1) = ln x Big|1^e. Substituting the limits gives ln(e) - ln(1). Since ln(e) = 1 and ln(1) = 0, the integral evaluates to 1 - 0 = 1.

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Evaluate a definite integral of 1/x using FTC Part 2

Use the Fundamental Theorem of Calculus, Part 2, to evaluate the definite integral:

$\int_1^e \frac{1}{x}\,dx$

What is the exact value of this integral?

Hint

Recall that an antiderivative of $\frac{1}{x}$ is $\ln|x|$ . Apply FTC Part 2 by evaluating this antiderivative at the upper and lower limits and subtracting: $F(e) - F(1)$ .

Step-by-step solution

Since $\frac{d}{dx}[\ln x] = \frac{1}{x}$ , an antiderivative of $f(x) = \frac{1}{x}$ is $F(x) = \ln x$ . By FTC Part 2, $\int_1^e \frac{1}{x}\,dx = F(e) - F(1) = \ln x \Big|_1^e$ .

Substituting the limits gives $\ln(e) - \ln(1)$ . Since $\ln(e) = 1$ and $\ln(1) = 0$ , the integral evaluates to $1 - 0 = 1$ .

Evaluate a definite integral of 1/x using FTC Part 2

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