Computing Definite Integrals — Question 9

Question 9 - Solution

Rewrite the integrand using exponents: $$\frac{2}{\sqrt{x}}-\frac{1}{x} = 2x^{-1/2}-x^{-1}.$$

Integrate term by term: $$\int 2x^{-1/2}\,dx = 2\cdot\frac{x^{1/2}}{1/2} = 4\sqrt{x},$$ $$\int x^{-1}\,dx = \ln x.$$

Thus, $$\int \left(\frac{2}{\sqrt{x}}-\frac{1}{x}\right)\,dx = 4\sqrt{x}-\ln x.$$

Evaluate from $1$ to $4$: $$\left[4\sqrt{x}-\ln x\right]_{1}^{4}.$$

Compute the upper bound: $$4\sqrt{4}-\ln 4 = 8-\ln 4.$$

Compute the lower bound: $$4\sqrt{1}-\ln 1 = 4.$$

Subtract: $$(8-\ln 4)-4 = 4-\ln 4.$$

$$\boxed{4-\ln 4}$$