Question

Verify the statement by showing that the derivative of the right side is equal to the integrand of the left side. $$ \int 2 \sqrt{x}(x-3) d x=\frac{4 x^{3 / 2}(x-5)}{5}+C $$

Short answer

The statement is indeed true as the derivative of the right side \( 2x^{3/2} - 6x^{1/2} \) is the same as the integrand on the left side \( 2x^{3/2} - 6x^{1/2} \).

Step-by-step solution

Understand the rules for differentiation

The Power rule states if f(x) = \( x^n \), then f'(x) = \( nx^{(n-1)} \). The Product rule states if \( f(x) = g(x)h(x) \), then \( f'(x) = g'(x)h(x) + g(x)h'(x) \). Both rules are necessary in proceeding with the problem.

Differentiation of the right side

Apply the product rule. First, consider \( f(x) = 4x^{3/2} \), differentiating yields \( f'(x) = 6x^{1/2} \). Next, consider \( g(x) = (x - 5) \), differentiating yields \( g'(x) = 1 \). Then differentiate the term \( \frac{f(x)g(x)}{5} = \frac{4x^{3/2} * (x - 5)}{5} \). Applying the product rule gives us \( \frac{f'(x) * g(x) + f(x) * g'(x)}{5} \) = \( \frac{6x^{1/2} * (x - 5) + 4x^{3/2} * 1}{5} \) = \( \frac{6x^{3/2} - 30x^{1/2} + 4x^{3/2}}{5} \) = \( 2x^{3/2} - 6x^{1/2} \). Finally, considering the constant C which differentiates to 0, we get \( 2x^{3/2} - 6x^{1/2} \) as the final derivative.

Compare with left side

Now compare the result of step 2 with the integral on the left. Simplification of the left integral gives \( 2x^{3/2} - 6x^{1/2} \). Both sides match confirming that the derivative of the right side is equal to the integrand of the left side.