Q.Evaluate $\int_{-1}^1 5x^4 \sqrt{x^5 + 1}\, dx$

Substitute $t = x^2 \Rightarrow dt = 2x\,dx$, and $x^3\,dx = x^2\cdot x\,dx = t\cdot\tfrac{1}{2}\,dt$. Limits: $x=0\to t=0$, $x=2\to t=4$.

Substitute again $s = 4-t \Rightarrow t = 4-s,\ dt = -ds$; limits flip $0\to4$:

Integrate.

Therefore

Let $x = s^3$, so $dx = 3s^2\,ds$; limits $x:0\to 8$ give $s:0\to 2$. Then $x^{5/3}=s^5$ and $x^{2/3}=s^2$:

Put $s = 2\sin\theta$, $ds = 2\cos\theta\,d\theta$, $4-s^2 = 4\cos^2\theta$, $s:0\to2 \Rightarrow \theta:0\to\tfrac{\pi}{2}$:

With the Beta/Wallis reduction,

Hence

The problem asks us to evaluate a definite integral. Integrals of this form, where the integrand involves a function $f(x)$ and $f(a+b-x)$ in a symmetric way, are often solved using a powerful property of definite integrals, commonly known as King's property.

Concept: King's Property for Definite Integrals

King's property states that for a continuous function $f(x)$ on the interval $[a, b]$:

Let's apply this concept to the given integral.

1. Identify the integral and its structure: Let the given integral be $I$:

The limits of integration are $a=5$ and $b=9$.
Let $f(x) = \log 3x^2$.
We need to check if the second term in the denominator, $\log(588 - 84x + 3x^2)$, can be expressed as $f(a+b-x)$.
First, calculate $a+b-x$:
$a+b-x = 5+9-x = 14-x$.
Now, let's find $f(a+b-x) = f(14-x)$:
$f(14-x) = \log 3(14-x)^2$.
Expand the argument of the logarithm:
$3(14-x)^2 = 3(14^2 - 2 \cdot 14 \cdot x + x^2)$
$3(14-x)^2 = 3(196 - 28x + x^2)$
$3(14-x)^2 = 588 - 84x + 3x^2$.
This exactly matches the argument of the second logarithm in the denominator of the original integral.
So, the integral is indeed of the form:

where $f(x) = \log 3x^2$, $a=5$, and $b=9$.

2. Apply King's Property:

Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, we replace $x$ with $a+b-x = 14-x$ in the integral $I$:

From our calculation in step 1, we know that $3(14-x)^2 = 588 - 84x + 3x^2$.
And similarly, $588 - 84(14-x) + 3(14-x)^2 = 3x^2$.
Substituting these back into the transformed integral:

Let's call the original integral (1) and this transformed integral (2):

3. Add the original and transformed integrals: Add equation (1) and equation (2):

Notice that the denominators of both fractions are identical. Let $D(x) = \log 3x^2 + \log(588 - 84x + 3x^2)$.

Since the numerator is also $D(x)$, the fraction simplifies to $1$:

4. Evaluate the simplified integral:

The value of the integral is $2$. This corresponds to option (A).

We start by evaluating the definite integral. The expression $\int_{0}^{3} (3x^2 - 4x + 2)\, dx$ is a straightforward polynomial integral. Once we find $k$, we plug it into the equation $3x^2 - 4x + 2 = \dfrac{3k}{5}$ and solve for $x$. The problem asks for an integer root, so we check which of the given options satisfies the equation.

1. Compute $k$ Integrate term by term:

Evaluate from $0$ to $3$:

So $k = 15$.

2. Set up the equation The problem gives:

Substitute $k = 15$:

Thus we solve:

3. Solve the quadratic Rearrange:

Factor or use the quadratic formula. Check factors:

Indeed, $(3x - 7)(x + 1) = 3x^2 + 3x - 7x - 7 = 3x^2 - 4x - 7$.

So the roots are:

4. Identify the integer root The problem asks for an integer root. Among the two solutions, $x = -1$ is an integer. The other root $\frac{7}{3}$ is not an integer. Checking the options, $-1$ appears as option (D).

The problem asks us to find the value of $a-b$ for a given cubic function $f(x) = x^3 + a x^2 + b x$, using two conditions. The core idea is to translate these conditions into algebraic equations involving $a$ and $b$, and then solve the resulting system of equations.

1. Understand the function and its derivative: The given function is $f(x) = x^3 + a x^2 + b x$. Since $f(x)$ is a polynomial, it is continuous and differentiable everywhere. We will need its derivative, $f'(x)$:

2. Use the first condition: $f(1) - f(3) = 0$ This condition means $f(1) = f(3)$. Let's calculate the values of $f(x)$ at $x=1$ and $x=3$:

Setting $f(1) = f(3)$:

Rearrange the terms to form a linear equation in $a$ and $b$:

Dividing the entire equation by 2 simplifies it:

> [!IMPORTANT] Rolle's Theorem
> If a function $f(x)$ is continuous on a closed interval $[p, q]$ and differentiable on the open interval $(p, q)$, and if $f(p) = f(q)$, then there exists at least one point $c \in (p, q)$ such that $f'(c) = 0$.
>
> In our case, $f(x)$ is a polynomial, so it's continuous on $[1, 3]$ and differentiable on $(1, 3)$. Since $f(1) = f(3)$, Rolle's Theorem guarantees that there must be some $c \in (1, 3)$ where $f'(c) = 0$. The second condition of the problem gives us exactly this value of $c$.

3. Use the second condition: $f'\left(\dfrac{2\sqrt{3} + 1}{\sqrt{3}}\right) = 0$

Let $c = \dfrac{2\sqrt{3} + 1}{\sqrt{3}}$. We can simplify this value:

We should verify that this value of $c$ lies in the interval $(1, 3)$. Since $\sqrt{3} \approx 1.732$, $\frac{1}{\sqrt{3}} \approx 0.577$. So, $c \approx 2 + 0.577 = 2.577$, which is indeed between 1 and 3.

Now, substitute $x = c = 2 + \frac{1}{\sqrt{3}}$ into the expression for $f'(x)$ and set it to zero:

First, let's expand the squared term:

To combine these, find a common denominator:

Substitute this back into the equation for $f'(c)$:

4. Solve the system of equations for $a$ and $b$

   We have two equations:

   1. $4a + b = -13 \implies b = -13 - 4a$
   2. $13 + 4\sqrt{3} + 4a + \frac{2a}{\sqrt{3}} + b = 0$

   Substitute the expression for $b$ from Equation 1 into Equation 2:

Notice the cancellations:

The $13$ and $-13$ cancel out. The $4a$ and $-4a$ cancel out.
This leaves us with a much simpler equation:

To solve for $a$, multiply the entire equation by $\sqrt{3}$:

Now that we have $a$, substitute it back into Equation 1 to find $b$:

So, we have $a = -6$ and $b = 11$.

5. Calculate $a-b$

Critical points in $[0,2]$: $x=\tfrac13$ and $x=1$.

Evaluate at the critical points and the endpoints:

Absolute maximum $M = -1$ (at $x=2$); absolute minimum $m = -3$ (at $x=0$ and $x=1$).

The problem asks us to evaluate the limit of a sum as $n$ approaches infinity. This is a classic scenario where the sum can be converted into a definite integral using the concept of Riemann sums.

Concept: Riemann Sums and Definite Integrals

A definite integral $\int_a^b f(x) dx$ represents the area under the curve of $f(x)$ from $x=a$ to $x=b$. This area can be approximated by summing the areas of many thin rectangles. As the number of rectangles approaches infinity (and their width approaches zero), the sum of their areas converges to the exact definite integral.

The standard form for converting a limit of a sum to a definite integral is:

The key is to identify $f(x)$ and the limits of integration by matching the terms in the given sum with the general form. We look for a term of the form $\frac{k}{n}$ or $a + \frac{k}{n}$ inside the function $f$.

Step-by-step Solution

1. Identify the general term of the sum: The given limit is:

Let's examine the arguments of the sine function:
*   First term: $\frac{\pi}{4}$
*   Second term: $\frac{\pi}{12} \left( 3 + \frac{1}{n} \right)$
*   Third term: $\frac{\pi}{12} \left( 3 + \frac{2}{n} \right)$
*   ...
*   Last term: $\frac{\pi}{3}$

We can rewrite the first term to fit the pattern:
$\frac{\pi}{4} = \frac{3\pi}{12} = \frac{\pi}{12} \times 3 = \frac{\pi}{12} \left( 3 + \frac{0}{n} \right)$.
Similarly, let's see if the last term fits the pattern for some $k$.
$\frac{\pi}{3} = \frac{4\pi}{12} = \frac{\pi}{12} \times 4 = \frac{\pi}{12} \left( 3 + 1 \right) = \frac{\pi}{12} \left( 3 + \frac{n}{n} \right)$.

So, the general term inside the sine function is $\frac{\pi}{12} \left( 3 + \frac{k}{n} \right)$, where $k$ ranges from $0$ to $n$.
The sum can be written as:

2. Convert the sum into a definite integral: We have the form $\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n} f\left(\frac{k}{n}\right)$. We can identify $x = \frac{k}{n}$ and $f(x) = \sin \left( \frac{\pi}{12} (3 + x) \right)$. The limits of integration will be from $x=0$ (when $k=0$) to $x=1$ (when $k=n$). Therefore, the limit of the sum is equal to the definite integral:

3. Evaluate the definite integral:

   To evaluate this integral, we use a substitution.

   Let $u = \frac{\pi}{12} (3 + x)$.

   Then, $du = \frac{\pi}{12} dx$, which means $dx = \frac{12}{\pi} du$.

   Now, we need to change the limits of integration according to the substitution:

   • When $x=0$, $u = \frac{\pi}{12} (3 + 0) = \frac{3\pi}{12} = \frac{\pi}{4}$.
   • When $x=1$, $u = \frac{\pi}{12} (3 + 1) = \frac{4\pi}{12} = \frac{\pi}{3}$.

   The integral becomes:

The integral of $\sin(u)$ is $-\cos(u)$.

Substitute the known values of cosine: $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

The calculated value matches option (B).

The concept first. A limit of the form $\frac{0}{0}$ means numerator and denominator share a common vanishing factor; the whole job is to expose it and cancel. Here both vanish at $x = 3/2$ (i.e. $2x = 3$), and the tool for a cube root difference is

applied with $a = u^{1/3},\; b = 3^{1/3}$, which converts the awkward $u^{1/3} - 3^{1/3}$ into the friendly factor $(u-3)$.

Step 1 — Substitute and factorise the numerator.

Let $u = 2x$; as $x \to \tfrac32$, $u \to 3$.

So

using $(u-3)(u^2+3u+9) = u^3 - 27$.