Problem 39 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Suppose that \(f^{\prime \prime}(x)\) is continuous for all \(x\) and \(f(0)=f^{\prime}(1)=1 .\) If \(\int_{0}^{1} t f^{\prime \prime}(t) d t=0\), then the value of \(f(1)\) is (A) 3 (B) 2 (C) \(4 \frac{1}{2}\) (D) None of these

Step-by-Step Solution

Answer

The value of \( f(1) \) is 2.

Step 1: Consider the Given Integral

We have the integral \( \int_{0}^{1} t f^{\prime \prime}(t) \, dt = 0 \). To solve this, use integration by parts. Let \( u = t \) and \( dv = f^{\prime \prime}(t) \, dt \).

Step 2: Apply Integration by Parts

Using integration by parts, where \( \int u \, dv = uv - \int v \, du \), set: \( u = t \), \( du = dt \), and \( v = f^{\prime}(t) \). So,\[ \int_{0}^{1} t f^{\prime \prime}(t) \, dt = \left[ t f^{\prime}(t) \right]_{0}^{1} - \int_{0}^{1} f^{\prime}(t) \, dt = 0. \]

Step 3: Evaluate the Boundary Terms

Evaluate \( \left[ t f^{\prime}(t) \right]_{0}^{1} \). At the upper limit, \( t = 1 \), so \( f^{\prime}(1) = 1 \). At the lower limit, \( t = 0 \), so the term equals zero. Thus, \( [1 \times 1] - 0 = 1 \).

Step 4: Solve the Remaining Integral Equation

Substitute the boundary term result into the integral equation:\[ 1 - \int_{0}^{1} f^{\prime}(t) \, dt = 0. \]Solve for \( \int_{0}^{1} f^{\prime}(t) \, dt \), giving \( \int_{0}^{1} f^{\prime}(t) \, dt = 1. \)

Step 5: Evaluate \( f(1) \) using Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, \( \int_{0}^{1} f^{\prime}(t) \, dt = f(1) - f(0) \). Since \( f(0) = 1 \), we have:\[ f(1) - 1 = 1. \]Thus, \( f(1) = 2. \)

Key Concepts

Integration by Parts

Integration by Parts is a powerful tool in Calculus used to integrate the product of two functions. This method is grounded in the differentiation rule for products and is especially useful when a direct integration method seems complicated. The formula for Integration by Parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

Here, you need to select which part of the integrand will be \( u \) and which part will be \( dv \). A good choice of \( u \) is a function that simplifies when differentiated, and \( dv \) should be easy to integrate.
In our exercise, we chose \( u = t \) and \( dv = f''(t) \; dt \). Differentiating \( u \) gives \( du = dt \), and integrating \( dv \) gives \( v = f'(t) \). This selection simplifies the integral effectively, revealing the relationship between the terms when solved by parts.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation with integration, serving as a backbone for the study of Calculus. It consists of two parts that explain the relationship between the definite integral and the antiderivative.

• First Part: Describes how the integral of a function's derivative over an interval gives the net change of the function over that interval.
• Second Part: If a function is continuous over an interval \([a, b]\), and \(F\) is its antiderivative, then \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\).

In our exercise, the theorem helps solve for \( f(1) \). We find \( \int_{0}^{1} f'(t) \; dt = f(1) - f(0) \), equating this to 1, as calculated in the solution. Knowing that \( f(0) = 1 \), we determine that \( f(1) = 2 \).
This essential concept shows the interplay between antiderivatives and integrals, reflecting the geometric interpretation of the area under curves.

Definite Integrals

Definite Integrals allow us to calculate the exact area under a curve from one point to another on the x-axis. The evaluation of a definite integral gives a number that represents this accumulated area, which can be positive, negative, or zero, depending on the curve’s position relative to the x-axis.
To evaluate a definite integral, you choose an antiderivative of the function, then apply the limits of integration according to the notation \([a, b]\). Proper utilization of integration techniques, such as Integration by Parts, can simplify this process.
In our given exercise, we used the definite integral \( \int_{0}^{1} t f''(t) \, dt = 0 \). This value hints that the total contribution from the product of \( t \) and \( f''(t) \) over the interval is zero.

• This informs the steps taken using Integration by Parts and ultimately helps to deduce values related to the function \( f(t) \).

Definite integrals are foundational in Calculus for solving practical problems involving areas, volumes, and total values over intervals.