Animated Solution for Mathematics - Definite Integration: Following question has matching lists. Match list I with List II.

We are tasked with finding the number of polynomials $f(x)=a{x}^2+bx+c$ with non-negative integer coefficients such that $f(0)=0$ and ${\int }_0^{1}f(x)dx=1$.

The condition $f(0)=0$ immediately implies that $c=0$. Thus, our polynomial simplifies to the form $f(x)=a{x}^2+bx$.

Integrating this expression from $0$ to $1$, we obtain:

Multiplying the entire equation by $6$, we arrive at the linear Diophantine equation:

Since $a$ and $b$ must be non-negative integers, we test the possible values for $b$:

1. If $b=0$, then $2a=6$, which gives $a=3$. 2. If $b=2$, then $3b=6$, which gives $2a=0$, so $a=0$.

Any other value for $b$ results in $a$ being either non-integer or negative. Therefore, there are exactly 2 such polynomials.

We examine the function $f(x)=\sin (x^2)+\cos (x^2)$ on the interval $[-\sqrt{13},\sqrt{13}]$. This expression follows the form $A\sin \theta +B\cos \theta$, which can be rewritten as:

The maximum value of this function is $\sqrt{2}$, which occurs when the argument of the sine function is $2n\pi +\frac{\pi }{2}$. Setting the argument accordingly:

Given the interval $[-\sqrt{13},\sqrt{13}]$, we have $x^2\in [0,13]$. We test values of $n$:

1. For $n=0$, $x^2=\frac{\pi }{4}\approx 0.78$, which yields two points: $x=\pm \sqrt{\frac{\pi }{4}}$. 2. For $n=1$, $x^2=2\pi +\frac{\pi }{4}=\frac{9\pi }{4}\approx 7.06$, which yields two more points: $x=\pm \sqrt{\frac{9\pi }{4}}$. 3. For $n=2$, $x^2=4\pi +\frac{\pi }{4}\approx 13.35$, which is outside the allowed range.

In total, there are 4 points where the function reaches its maximum.

We evaluate the integral $I={\int }_{-2}^2\frac{3x^2}{1+e^x}dx$. We apply King's property, which states that ${\int }_{-a}^{a}f(x)dx={\int }_{-a}^{a}f(-x)dx$.

Replacing $x$ with $-x$, we get:

Adding the two versions of $I$ together:

Evaluating this integral:

Thus, the final value is $I=8$.

Finally, we consider the integral of $g(x)=\cos (2x)\log \left(\frac{1+x}{1-x}\right)$. We check the parity of the function by evaluating $g(-x)$:

Since $g(-x)=-g(x)$, the function is odd. The integral of any odd function over a symmetric interval $[-a,a]$ is identically zero.

Consequently, the numerator of the ratio is zero, making the final result 0.