Problem 77 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. \(\int_{0}^{1} x \sqrt{x+1} d x=\sqrt{x+1} \int_{0}^{1} x d x=\left.\frac{1}{2} x^{2} \sqrt{x+1}\right|_{0} ^{1}=\frac{\sqrt{2}}{2}\)78. If \(f^{\prime}\) is continuous on \([0,2]\), then \(\int_{0}^{2} f^{\prime}(x) d x=f(2)-f(0)\).

Step-by-Step Solution

Answer

The given statement is partly false and partly true. The integral \(\int_{0}^{1} x \sqrt{x+1} d x\), when evaluated correctly, does not simplify to \(\frac{\sqrt{2}}{2}\). The method provided in the statement for separating the integral into two parts is incorrect. However, the second part of the statement which claims that if \(f'\) is continuous on \([0,2]\), then \(\int_{0}^{2} f'(x) dx = f(2) - f(0)\) is true, as it is a consequence of the Fundamental Theorem of Calculus.

Step 1: Understand the main claim

The main claim in the statement is that $\int_{0}^{1} x \sqrt{x+1} d x=\sqrt{x+1} \int_{0}^{1} x d x=\left.\frac{1}{2} x^{2} \sqrt{x+1}\right|_{0} ^{1}=\frac{\sqrt{2}}{2}$. We will evaluate it step-by-step.

Step 2: Check the initial manipulation of the integral

The statement claims that we can separate the integral into two parts: \(\sqrt{x+1}\) and \(\int_{0}^{1} x dx\). This is incorrect, as it would mean that the given expression is a product of two functions and we could use a multiplication rule for integration. However, such a rule does not exist in calculus. Therefore, the statement is false.

Step 3: Evaluate the given integral correctly

To double-check that the statement is wrong, let's evaluate the given integral correctly. We have: \(\int_{0}^{1} x \sqrt{x+1} d x\). We can use substitution: let \(u = x+1\), then \(x = u-1\) and \(dx = du\). Now, we change the bounds of the integral: for \(x=0\), \(u=1\); for \(x=1\), \(u=2\). The new integral is: \(\int_{1}^{2} (u-1) \sqrt{u} du\). Now we can integrate by expanding the expression inside the integral: \(\int_{1}^{2} (u^{3/2}-u^{1/2}) du = \left.\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right|_{1} ^{2}\). Evaluating it at the bounds, we get: \(\left(\frac{2}{5}(2)^{5/2} - \frac{2}{3}(2)^{3/2}\right) - \left(\frac{2}{5}(1)^{5/2} - \frac{2}{3}(1)^{3/2}\right)\). This simplifies to a different value than \(\frac{\sqrt{2}}{2}\).

Step 4: Evaluate the second part of the statement

The second part of the statement claims that if \(f'\) is continuous on \([0,2]\), then \(\int_{0}^{2} f'(x) dx = f(2) - f(0)\). This is actually true. It is a consequence of the Fundamental Theorem of Calculus: if \(F\) is an antiderivative of a continuous function \(f\) on an interval \([a, b]\), then \(\int_{a}^{b} f(x) dx = F(b) - F(a)\). In this case, \(f'(x)\) is continuous, so the theorem is applicable. We can write \(f'(x) = f\), then \(f(x)\) is an antiderivative of \(f'(x)\). Therefore, \(\int_{0}^{2} f'(x) dx = f(2) - f(0)\) is true. #Conclusion# The original statement has a false part and a true part. The initial solution for the integral is incorrect, which results in a false conclusion for that part. However, the second part of the statement about the continuity of \(f'\) and its integral is true, coming from the Fundamental Theorem of Calculus.