Question

A function \(\mathrm{f}\) is defined for all real \(\mathrm{x}\) by the formula \(\mathrm{f}(\mathrm{x})=3+\int_{0}^{\mathrm{x}} \frac{1+\sin \mathrm{t}}{2+\mathrm{t}^{2}} \mathrm{dt}\). Without attempting to evaluate this integral, find a quadratic polynomial \(\mathrm{p}(\mathrm{x})=\mathrm{a}+\mathrm{bx}+\mathrm{cx}^{2}\) such that \(\mathrm{p}(0)=\mathrm{f}(0), \mathrm{p}^{\prime}(0)=\mathrm{f}^{\prime}(0)\), and \(\mathrm{p}^{\prime \prime}(0)=\) f' \((0)\).

Short answer

Question: Find the quadratic polynomial p(x) = a + bx + cx^2 such that p(0) = f(0), p'(0) = f'(0), and p''(0) = f''(0) for the function \(f(x) = 3 + \int_0^x \frac{1+\sin t}{2+t^2}dt\). Answer: The quadratic polynomial is \(p(x) = 3 + \frac{1}{2}x + \frac{1}{4}x^2\).

Step-by-step solution

Find f(0)

To find f(0), plug x = 0 into the given formula: \(f(0) = 3 + \int_0^0 \frac{1+\sin t}{2+t^2}dt\) Since the integral is from 0 to 0, the integral is zero: \(f(0) = 3 + 0 = 3\)

Find f'(x) using the Fundamental Theorem of Calculus

The derivative of the function f(x) can be found using the Fundamental Theorem of Calculus: \(f'(x) = \frac{d}{dx}(3 + \int_0^x \frac{1+\sin t}{2+t^2}dt) = \frac{1+\sin x}{2+x^2}\)

Find f'(0) and f''(x)

To find f'(0), plug x = 0 into f'(x): \(f'(0) = \frac{1+\sin 0}{2+0^2} = \frac{1}{2}\) To find f''(x), take the derivative of f'(x) with respect to x: \(f''(x) = \frac{d}{dx}(\frac{1+\sin x}{2+x^2})\) Using the quotient rule: \(f''(x) = \frac{(2+x^2)\cos x - (1+\sin x)(2x)}{(2+x^2)^2}\) Now find f''(0): \(f''(0) =\frac{(2+0^2)\cos 0 - (1+\sin 0)(0)}{(2+0^2)^2} = \frac{2}{4} =\frac{1}{2}\)

Determine the coefficients of p(x)

We are given the polynomial p(x) = a + bx + cx^2 with the conditions: 1. p(0) = f(0) = 3 2. p'(0) = f'(0) = 1/2 3. p''(0) = f''(0) = 1/2 Since p(0) = a, we have a = 3. To find the coefficients b and c, we need to find the derivatives of p(x): \(p'(x) = b + 2cx\) \(p''(x) = 2c\) Given p'(0)=f'(0)=1/2, set x=0 and solve for b: \(p'(0) = b + 2c(0) = b\) Therefore, b = 1/2. Given p''(0)=f''(0)=1/2, we can solve for c: \(p''(0) = 2c\) Therefore, c = 1/4.

Write the quadratic polynomial p(x)

Using the coefficients a, b, and c we found in the previous steps, we can write the quadratic polynomial p(x) as: \(p(x) = a + bx + cx^2 = 3 + \frac{1}{2}x + \frac{1}{4}x^2\)

Key concepts

Fundamental Theorem of Calculus

Understanding the Fundamental Theorem of Calculus (FTC) is crucial for anyone delving into integral calculus. The theorem bridges the concepts of differentiation and integration, two fundamental operations in calculus.

Essentially, the FTC states that if a function is continuous on an interval and we define a function, F(x), as the integral of this function from a fixed start point to x, then F(x) is also differentiable on that interval, and its derivative at any point is equal to the original function. Symbolically, given a continuous function f(t), if we define F(x) = \[ \int_{a}^{x} f(t) \,dt \], then the derivative of this integrated function is F'(x) = f(x).

This fundamental insight not only helps in solving definite integrals but also in finding a function that represents the accumulation of areas under a curve. In the context of our exercise, the FTC helps in finding the derivative of the function f(x), without having to evaluate the integral directly.

Quadratic Polynomial

A quadratic polynomial is an expression of the form \( p(x) = ax^2 + bx + c \), where a, b, and c are constants, and 'a' is not equal to zero. These polynomials are second-degree polynomials, meaning the highest power of x is two.

The graph of a quadratic polynomial is a parabola, which can open upwards or downwards based on the sign of the coefficient 'a'. When 'a' is positive, the parabola opens upwards, and when 'a' is negative, it opens downwards. Quadratics have various properties, including a vertex, axis of symmetry, and they can have either zero, one, or two real roots.

In our solved exercise, we were looking to match a quadratic polynomial to certain conditions given by the function f(x). The coefficients of this polynomial possess the values determined from f(x) and its derivatives, which illustrates an application of both calculus and algebra in solving the problem.

Derivative and Integration Techniques

While solving calculus problems, one often employs various derivative and integration techniques. Differentiation involves finding the rate at which a function is changing at any given point, which is represented by the derivative. The basic techniques include using the power rule, product rule, quotient rule, and chain rule.

In the exercise we reviewed, the quotient rule is used, which applies when you have a function divided by another function. The quotient rule states: \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \), where u and v are differentiable functions of x.

On the other side, integration techniques are ways to find the antiderivative or integral of a function. Techniques like substitution, integration by parts, partial fractions, and trigonometric integration are common. These techniques are essential for solving more complex integrals that do not resemble standard forms.

Mastery of both derivative and integration techniques is essential for any calculus student, particularly when solving problems pertaining to the IIT JEE where a strong conceptual understanding and skillset are required.