Computing Definite Integrals — Question 7

Question 7

Evaluate the definite integral $$\int_{0}^{1} (3x^2+2)\sqrt{x}\,dx.$$

Question 7 - Solution

Begin by rewriting the integrand using powers of $x$: $$(3x^2+2)\sqrt{x} = 3x^{5/2}+2x^{1/2}.$$

Now integrate term by term: $$\int 3x^{5/2}\,dx = 3\cdot\frac{x^{7/2}}{7/2} = \frac{6}{7}x^{7/2},$$ $$\int 2x^{1/2}\,dx = 2\cdot\frac{x^{3/2}}{3/2} = \frac{4}{3}x^{3/2}.$$

Combine the results: $$\int (3x^2+2)\sqrt{x}\,dx = \frac{6}{7}x^{7/2}+\frac{4}{3}x^{3/2}.$$

Evaluate from $0$ to $1$: $$\left[\frac{6}{7}x^{7/2}+\frac{4}{3}x^{3/2}\right]_{0}^{1} = \frac{6}{7}+\frac{4}{3}.$$

Find a common denominator: $$\frac{6}{7}+\frac{4}{3} = \frac{18+28}{21} = \frac{46}{21}.$$

$$\boxed{\frac{46}{21}}$$