Evaluate: , if it exists. | Basic Mathematics 2081

Basic Mathematics20815 marksNumerical: improper integral evaluationImproper integrals

Evaluate: $\int_0^5 \frac{dx}{\sqrt{x - 2}}$ , if it exists. [5]

Evaluate: $$\int_0^5 \frac{dx}{\sqrt{x - 2}}$$, if it exists. [5]

Integrand: $f(x) = \dfrac{1}{\sqrt{x-2}}$ - Lower limit: $0$ - Upper limit: $5$ The integrand $\dfrac{1}{\sqrt{x-2}}$ is only defined (in real numbers) when $x - 2 0$ , i.e. $x 2$ . For $x < 2$ , the quantity $x - 2 < 0$ , so $\sqrt{x-2}$ is not a real number. The interval of integration is

Integrand: $f(x) = \dfrac{1}{\sqrt{x-2}}$ - Lower limit: $0$ - Upper limit: $5$ The integrand $\dfrac{1}{\sqrt{x-2}}$ is only defined (in real numbers) when $x - 2 0$, i.e. $x 2$. For $x < 2$, the quantity $x - 2 < 0$, so $\sqrt{x-2}$ is not a real number. The interval of integration is