1. The problem states two functions. First, \( f(x) = \sqrt{x+1} - 1 \) with its inverse \( g(x) = x+1 \). 2. The domain of \( f \) is \( x \geq -1 \) because the expression under the square root \( x+1 \) must be non-negative. 3. To verify that \( g \) is the inverse of \( f \), check the compositions: $$ f(g(x)) = f(x+1) = \sqrt{(x+1)+1} - 1 = \sqrt{x+2} - 1, $$ which simplifies correctly based on domain restrictions. $$ g(f(x)) = g(\sqrt{x+1} -1) = (\sqrt{x+1} -1) +1 = \sqrt{x+1}. $$ Given the condition \( f \circ g(x) = x \) holds in domain, confirm \( f^{-1} = g \). 4. Next, evaluate the integral \( \int_0^1 f(x) \, dx = \int_0^1 \left(\sqrt{x+1} -1 \right) dx \). 5. Calculate: $$ \int_0^1 \sqrt{x+1} \, dx = \int_1^2 \sqrt{u} \, du \quad \text{with } u = x+1, du=dx. $$ 6. Integrate: $$ \int_1^2 u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{2}{3} (2^{3/2} - 1). $$ 7. Also, \( \int_0^1 1 \, dx = 1. \) 8. So, $$ \int_0^1 f(x) \, dx = \frac{2}{3} (2^{3/2} - 1) - 1 = \frac{2}{3} (2\sqrt{2} - 1) - 1. $$ 9. For the second problem, consider \( f(x) = x^3 + 2x - 3 \) and \( (x+1)^2 - 4 \). 10. To show the relation, observe that $$ (x+1)^2 - 4 = x^2 + 2x + 1 - 4 = x^2 + 2x - 3. $$ 11. This expression resembles related forms but is not the same as \( f(x) \). Further details or clarification might be needed. 12. Graphs can be sketched to display the shapes: \( f(x) = x^3 + 2x - 3 \) is a cubic curve, and \( (x+1)^2 - 4 \) is a parabola shifted left and down. Final answers: - \( f(x) = \sqrt{x+1} - 1 \) with inverse \( g(x) = x + 1 \). - Integral \( \int_0^1 f(x) \, dx = \frac{2}{3} (2\sqrt{2} - 1) - 1 \). - The second set involves comparing cubic and quadratic curves.