Question

Show that \(\int_{0}^{1} x f^{\prime \prime}(x) d x=3\) for every function \(f(x)\) that satisfies the following conditions : (i) \(\mathrm{f}(\mathrm{x})\) is defined for all \(\mathrm{x}\), (ii) \(\mathrm{f}^{\prime \prime}(\mathrm{x})\) is continuous, (iii) \(f(0)=f(1)\), (iv) \(f^{\prime}(1)=3\)

Short answer

Question: Prove that if a function \(f(x)\) satisfies the conditions (i) \(f\) is defined for all \(x\); (ii) \(f^{\prime \prime}(x)\) is continuous; (iii) \(f(0)=f(1)\); and (iv) \(f^{\prime}(1)=3\), then \(\int_{0}^{1} x f^{\prime \prime}(x) dx = 3\). Answer: Apply integration by parts and evaluate the limits and integrals, resulting in \(\int_{0}^{1} x f^{\prime \prime}(x) dx = 3\).

Step-by-step solution

Apply Integration by parts

First, we apply integration by parts formula: $$\int u dv = uv - \int v du$$ We choose \(u=x\) and \(dv=f^{\prime \prime}(x)dx\). Now, we find \(du\) and \(v\). $$du = dx$$ To find \(v\), we need to find the antiderivative of \(f^{\prime \prime}(x)dx\): $$v = \int f^{\prime \prime}(x) dx = f^{\prime}(x) + C$$ Now, let's substitute \(u, v, du\) and \(dv\) in the integration by parts formula: $$\int_{0}^{1} x f^{\prime \prime}(x) dx = \left[x(f^{\prime}(x)+C)\right]_0^1 - \int_{0}^{1} (f^{\prime}(x)+C) dx$$

Evaluate the limits and integral

Now, we need to evaluate the limits and the integral: When \(x=1\): $$\left[x (f^{\prime}(x)+C)\right]_1 = 1(f^{\prime}(1)+C)$$ We know that \(f^{\prime}(1)=3\), so $$1(f^{\prime}(1)+C)=3+C$$ When \(x=0\): $$\left[x (f^{\prime}(x)+C)\right]_0 = 0(f^{\prime}(0)+C) = 0$$ Now, we evaluate the integral: $$\int_{0}^{1} (f^{\prime}(x)+C) dx = \left[\int_{0}^{1} f^\prime(x)dx\right] + C\left[\int_{0}^{1} dx\right]$$ To solve this integral, we find the antiderivative of \(f^\prime(x)\) which is \(f(x)\): $$\left[\int_{0}^{1} f^\prime(x)dx\right] = \left[f(x)\right]_0^1 = f(1)-f(0)$$ We know that \(f(0)=f(1)\), which means \(f(1)-f(0)=0\). $$C\left[\int_{0}^{1} dx\right] = C\left[x\right]_0^1 = C(1-0)=C$$ Now, we can substitute our results back into the original equation: $$\int_{0}^{1} x f^{\prime \prime}(x) dx = (3+C) - 0 - (0+C) = 3$$ Thus, we have shown that \(\int_{0}^{1} x f^{\prime \prime}(x) dx = 3\) for every function \(f(x)\) that satisfies the given conditions.

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the area beneath curves. It's best known for its operations that reverse the differentiation process, which is called finding the antiderivative or integral.

Consider the function that relates the rate of change (the derivative) to the accumulation of change, the integral highlights the total sum of these changes over an interval. For example, if a function represents the speed of a car at any given moment, the integral of that function from one time to another tells us the distance travelled.

In the exercise provided, integral calculus is used to find the value of an integral under specific conditions imposed on a function. The properties of integral calculus, such as the linearity of integrals and the fundamental theorem of calculus, play a crucial role in computing the solution to the problem.

Continuous Functions

Continuous functions are mathematical expressions that do not have sudden jumps or breaks. They can be graphically represented as a single unbroken curve. In the context of integral calculus, continuous functions are critical as they ensure that the integrals are well-defined.

Intuitively, for a function to be continuous at a point, you should be able to draw its graph at that point without lifting your pen off the paper. In our exercise, the given function has a second derivative that is continuous, which is a condition that ensures the applicability of numerous calculus operations, including integration by parts.

Importance of Continuity in Integration

When dealing with integrals, the continuity of a function or its derivatives ensures that the area under the curve is well-behaved and can be accurately calculated. If the second derivative wasn't continuous, the process of finding an antiderivative might lead to undefined or inaccurate results.

Antiderivatives

Antiderivatives are the opposite of derivatives in calculus. While a derivative represents the rate of change of a function, an antiderivative represents the original function before it was differentiated. Finding an antiderivative means discovering a function whose derivative is the given function.

This process is central to solving integration problems, where you often need to find an antiderivative to evaluate a definite integral. In the context of our exercise, integration by parts requires the finding of an antiderivative for the second derivative of the function. One can think of antiderivatives as the pieces to a puzzle that, when correctly assembled, reveal the bigger picture of the integral's value.

Applying Antiderivatives

Just as with the function in our exercise, when given the derivative of a function (like the second derivative), the aim is to trace back to the original function. This reversal process is a cornerstone of integral calculus as it allows us to compute areas, solve differential equations, and much more. In solving by integration by parts, understanding and applying antiderivatives is essential to reach the final solution.