Question

If \(\mathrm{f}(\mathrm{x})\) is a non-negative continuous function such that \(\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}+1 / 2)=1\), then find the value of \(\int_{0}^{2} f(x) d x\)

Short answer

Answer: The value of the integral ∫₀² f(x) dx is 1.

Step-by-step solution

Express f(x) using the given condition

Since we have the condition \(f(x) + f(x + \frac{1}{2}) = 1\), we can express f(x) as \(f(x) = 1 - f(x + \frac{1}{2})\).

Substitute f(x) in the integral

Now we need to find the value of \(\int_{0}^{2} f(x) dx\). Using the expression we found in step 1, the integral becomes: \(\int_{0}^{2} (1 - f(x + \frac{1}{2})) dx\).

Use substitution method for integration

Let's make the substitution \(u = x + \frac{1}{2}\). Then, \(du = dx\) and when \(x = 0\), \(u = \frac{1}{2}\) and when \(x = 2\), \(u = \frac{5}{2}\). Therefore, the integral becomes: \(\int_{\frac{1}{2}}^{\frac{5}{2}} (1 - f(u)) du\).

Split the integral

Now, we split the integral into two parts and solve them separately: $$\int_{\frac{1}{2}}^{\frac{5}{2}} (1 - f(u)) du = \int_{\frac{1}{2}}^{\frac{5}{2}} 1 du - \int_{\frac{1}{2}}^{\frac{5}{2}} f(u) du.$$

Evaluate the first integral

The first integral is quite simple: $$\int_{\frac{1}{2}}^{\frac{5}{2}} 1 du = u \Big|_{\frac{1}{2}}^{\frac{5}{2}} = \frac{5}{2} - \frac{1}{2} = 2.$$

Recognize that the second integral is same as the original integral

Observe that the second integral is equivalent to the original integral, but with different limits of integration: $$\int_{\frac{1}{2}}^{\frac{5}{2}} f(u) du = \int_{0}^{2} f(x) dx.$$ Therefore, we can write the entire equation as: $$\int_{0}^{2} f(x) dx = 2 - \int_{0}^{2} f(x) dx.$$

Solve for the original integral

To find the value of \(\int_{0}^{2} f(x) dx\), we can now add the integral to both sides of the equation: $$2 \int_{0}^{2} f(x) dx = 2.$$ Finally, divide both sides by 2: $$\int_{0}^{2} f(x) dx = 1.$$ The value of the integral \(\int_{0}^{2} f(x) dx\) is 1.