Find a polynomial with specified values given by an integral equation

Let

$f(x) = 3 + \int_0^x \frac{1+ \sin t}{2 + t^2} \, dt \qquad \text{for all } x \in \mathbb{R}.$

Find a quadratic polynomial $p(x) = a + bx + cx^2$ such that

$p(0) = f(0), \quad p'(0) = f'(0), \quad \text{and} \quad p''(0) = f''(0).$

Do not attempt to evaluate the integral.

First,

$f(0) = 3 + \int_0^0 \frac{1+\sin t}{2 + t^2} \, dt = 3.$

Hence,

$p(0) = 3 \quad \implies \quad a = 3.$

Next, by the fundamental theorem of calculus,

$f'(x) = \frac{1+\sin x}{2 + x^2} \quad \implies \quad f'(0) = \frac{1}{2}.$

Then, evaluate the derivative of the polynomial,

$p'(x) = b + 2cx \quad \implies \quad p'(0) = \frac{1}{2} = b.$

To get the final constant we take the derivative of $f'(x)$ ,

$\begin{align*} f''(x) &= \frac{ (\cos x)(2+x^2) - (1+\sin x)(2x)}{(2+x^2)^2} \\ \implies \quad f''(0) &= \frac{1}{2}. \end{align*}$

Therefore,

$p''(x) = 2c \quad \implies \quad p''(0) = \frac{1}{2} = 2c \quad \implies \quad c = \frac{1}{4}.$

Hence, we have

$p(x) = 3 + \frac{1}{2} x + \frac{1}{4} x^2.$

Stumbling Robot