Question

Let fhave derivatives of all orders. (a) Explain why \(f(b)=f(0)+\int_{0}^{b} f^{\prime}(x) d x\). (b) Using an integration by parts on the derivative integral in (a), with \(\mathrm{u}=\mathrm{f}^{\prime}(\mathrm{x})\) and \(\int \mathrm{vdx}=\mathrm{x}-\mathrm{b}\), show that \(f(\mathrm{~b})=\mathrm{f}(0)+\mathrm{f}^{\prime}(0) \mathrm{b}+\) \(\int_{0}^{b} f^{(2)}(x)(b-x) d x\) (c) Similarly, show that \(\mathrm{f}(\mathrm{b})=\mathrm{f}(0)+\mathrm{f}^{\prime}(0) \mathrm{b}+\frac{\mathrm{f}^{(2)}(0)}{2} \mathrm{~b}^{2}\) \(+\frac{1}{2} \int_{0}^{b} f^{(3)}(x)(b-x)^{2} d x\) (d) Check that (c) is correct for any quadratic polynomial. (e) Use another integration by parts on the formula in (c) to obtain the next formula.

Short answer

#tag_title#Answer:#tag_content# \(f(b) = f(0) + f'(0)b +\frac{1}{2}f^{(2)}(0)b^2 -\frac{1}{6}f^{(3)}(0)b^3+ \frac{1}{6}\int_{0}^{b} f^{(4)}(x)(3b^2x-3bx^2+x^3) dx\)

Step-by-step solution

Part (a):

First, we want to explain why \(f(b) = f(0) + \int_{0}^{b} f'(x) dx\). By the Fundamental Theorem of Calculus, we know that: \(f(b) = f(0) + \int_{0}^{b} f'(x) dx\) The theorem states that the integral of a function's derivative over an interval is equal to the change in the function's values over that interval. In this case, we are looking at the change from x=0 to x=b, so we have \(f(b)-f(0)\) as our change. This is equal to the integral of the derivative, so the result holds as stated.

Part (b):

Applying integration by parts to the integral in part (a), we are given u = f'(x) and \(v = \int (x-b) dx\). First, we need to differentiate u and integrate v to calculate du and dv: du = f''(x) dx \(v = \int (x-b) dx = \frac{1}{2}(x^2 - 2bx + b^2)\) Now, using the integration by parts formula, which states that: \(\int u \, dv = uv - \int v \, du,\) we obtain: \(f(b) = f(0) + f'(0)b - \int_{0}^{b} \frac{1}{2}(x^2 - 2bx + b^2) f''(x) dx\) Rearranging the terms gives: \(f(b) = f(0) + f'(0)b +\int_{0}^{b} f^{(2)}(x)(b-x) dx\)

Part (c):

By applying integration by parts once more to the integral part in the expression derived in part (b), and using u = f''(x), we get: \(v = \int (b-x)\,dx = \frac{1}{2}(x^2 - 2bx + b^2)\) Using the integration by parts formula, we have: \(f(b) = f(0) + f'(0)b +\frac{1}{2}f^{(2)}(0)b^2 - \int_{0}^{b} \frac{1}{2}(x^2 - 2bx + b^2)f'''(x) dx\) Rearranging the terms, we get: \(f(b) = f(0) + f'(0)b +\frac{1}{2}f^{(2)}(0)b^2 + \frac{1}{2}\int_{0}^{b} f^{(3)}(x)(b-x)^2 dx\)

Part (d):

To check that the expression in part (c) is correct for any quadratic polynomial, let: \(f(x) = ax^2 + bx + c\) Now, calculating the derivatives of f(x): \(f'(x) = 2ax + b\) \(f''(x) = 2a\) \(f'''(x) = 0\) Observe that since f'''(x) = 0, the integral term in part (c) will become zero as well. Thus, we have: \(f(b) = f(0) + f'(0)b +\frac{1}{2}f^{(2)}(0)b^2\) Now substituting the derivatives, we get \(f(b) = c + b*b +\frac{1}{2}(2a)*b^2 = a*b^2 + b*b + c\) which is exactly equal to f(x) evaluated at x=b, hence verifying the result for any quadratic polynomial.

Part (e):

Using the formula obtained in part (c), and performing another integration by parts, we have: \(u = f^{(3)}(x)\) \(v = \int (b - x)^2 dx = \frac{1}{3}(x^3 - 3bx^2 + 3b^2x)\) Now, applying the integration by parts formula, we get: \(f(b) = f(0) + f'(0)b +\frac{1}{2}f^{(2)}(0)b^2 + \frac{1}{2}f^{(3)}(0)(-b^3) - \frac{1}{2}\int_{0}^{b}\frac{1}{3}(x^3 - 3bx^2 + 3b^2x)f^{(4)}(x)dx\) Rearranging the terms, we have: \(f(b) = f(0) + f'(0)b+\frac{1}{2}f^{(2)}(0)b^2 -\frac{1}{6}f^{(3)}(0)b^3+ \frac{1}{6}\int_{0}^{b} f^{(4)}(x)(3b^2x-3bx^2+x^3)dx\)

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of antiderivatives with the area under the curve, providing one of the most powerful tools in calculus. It essentially states that if a function is continuous on a closed interval \[a, b\], then the integral of its derivative over that interval equals the change in its values at the endpoints of that interval. In mathematical terms, for a continuous function \(f\), we have \(f(b) - f(a) = \int_{a}^{b} f'(x) dx\).

This principle not only helps us evaluate definite integrals without the need for anti-differentiation but also plays a key role in understanding motion and area problems in calculus. For example, if you have the rate of change of a quantity, you can use this theorem to find the total change of the quantity over a certain time period. In the context of our exercise, where \(f\) has derivatives of all orders, it guarantees that \(f(b) = f(0) + \int_{0}^{b} f'(x) dx\), emphasizing the accumulative nature of the integral.

Integration by Parts

Integration by Parts is a technique derived from the product rule for differentiation and is used to integrate the product of two functions. The formula for Integration by Parts is \( \int u dv = uv - \int v du\). The choice of \(u\) and \(dv\) is critical and often follows the LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) rule or simply by choosing \(u\) to be the function that becomes simpler when differentiated.

In the IIT JEE exam, mastering this technique is crucial for tackling complex integrals. When we're given a product of functions to integrate, we use this method to transform the integral into a form that's easier to solve. For the given exercise, integration by parts allows us to express the integral \( \int_{0}^{b} f'(x) dx\) in terms of \( f'(0) \), \( f^{(2)}(0) \), and higher-order derivatives of \( f \). As the question progresses, integration by parts adeptly unwraps each layer of the function's derivatives, demonstrating its usefulness in handling intricate integrations.

Higher Order Derivatives

Higher Order Derivatives refer to derivatives of a function taken two or more times successively. The notation \( f^{(n)}(x) \) is commonly used to represent the n-th derivative of \( f \) with respect to \( x \). While the first derivative represents the rate of change of the function, higher-order derivatives can provide insights into the curvature and other aspects of the function's geometry.

Understanding higher-order derivatives is imperative for IIT JEE aspirants as they pertain directly to concepts like acceleration (which is the second derivative of position with respect to time) and can be used to solve problems involving series expansion, like Taylor and Maclaurin series. In our exercise, higher-order derivatives \( f^{(2)}(x) \) and \( f^{(3)}(x) \) are incorporated into the integration process, hinting at the complexity and dynamics of the function beyond its immediate rate of change. This elevates the problem-solving strategy and showcases the layered nature of calculus concepts in analyzing functions.

Quadratic Polynomials Integration

Integration of quadratic polynomials involves finding the antiderivative of a function of the form \( ax^2 + bx + c \). This is a fundamental skill for students preparing for the IIT JEE, as quadratics are a staple in calculus problems. When integrating a quadratic polynomial, we can straightforwardly apply the power rule of integration, which states that \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), for any real number \( n \) not equal to -1.

In the context of the exercise, verifying part (c) with a quadratic polynomial serves as a practical example of how higher-order derivatives of a polynomial function eventually diminish to zero. Since the third derivative of any quadratic polynomial is zero, the complexity added by the integral of the third derivative in previous steps disappears. This demonstrates the ease of integration when dealing with polynomial functions and provides students with a solid example of how calculus concepts apply to different types of functions.