Question

Compute the Riemann integral of the function f (x) = x on the interval [−1, 1].

Answer 1

Answer on Question #64865 – Math – Real Analysis

Question

Compute the Riemann integral of the function $f(x) = x$ on the interval $[-1, 1]$.

Solution

Theorem 1. If a function $f$ is continuous on an interval $[a, b]$, it is also Riemann-integrable on this interval.

Note that $f(x) = x$ is elementary function, then it is continuous on its domain, so $f(x) = x$ is integrable on the interval $[-1, 1]$ and integral $\int_{-1}^{1} f(x) dx$ exists.

Compute this integral.

Method 1

Function is odd if $f(-x) = -f(x)$. Note that $f(-x) = -x = -f(x)$, hence $f(x) = x$ is odd.

Theorem 2. Let the real function $f$ be Riemann-integrable on $[-a, a]$ and if $f$ is an odd function, then $\int_{-a}^{a} f(x) dx = 0$.

By Theorem 2 we get $\int_{-1}^{1} f(x) dx = 0$.

Method 2

By the Table of Integrals

$$\int x dx = \frac{x^2}{2} + C$$

and

Newton-Leibniz rule

$$\int_{a}^{b} f(x) dx = F(b) - F(a) = F(x) \Big|_{a}^{b}, \text{ where } F' = f,$$

we get

$$\int_{-1}^{1} f(x) dx = \int_{-1}^{1} x dx = \frac{x^2}{2} \Big|_{-1}^{1} = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0.$$

Answer: 0.

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