Key Concepts and Formulas

• Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, for $n \neq -1$.
• Integral of $\sqrt{a^2-x^2}$: $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.
• Geometric Interpretation of Integrals: Definite integrals can represent areas under curves.
• Substitution Rule: If $u = g(x)$, then $\int f(g(x))g'(x) \, dx = \int f(u) \, du$.

Step-by-Step Solution

Step 1: Evaluate the Left-Hand Side Integral

We need to evaluate $LHS = \int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}$. Let's split this into two integrals: $LHS = \int\limits_0^2 {\sqrt {2x} \, dx} - \int\limits_0^2 {\sqrt {2x - {x^2}} \, dx}$

For the first integral, $\int\limits_0^2 {\sqrt {2x} \, dx}$: $\int\limits_0^2 {\sqrt {2} \sqrt{x} \, dx} = \sqrt{2} \int\limits_0^2 {x^{1/2} \, dx} = \sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_0^2 = \sqrt{2} \left[ \frac{2}{3} x^{3/2} \right]_0^2$ $= \frac{2\sqrt{2}}{3} (2^{3/2} - 0^{3/2}) = \frac{2\sqrt{2}}{3} (2\sqrt{2}) = \frac{8}{3}$

For the second integral, $\int\limits_0^2 {\sqrt {2x - {x^2}} \, dx}$: We complete the square inside the square root: $2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -( (x-1)^2 - 1 ) = 1 - (x-1)^2$. So, the integral becomes $\int\limits_0^2 {\sqrt {1 - (x-1)^2} \, dx}$. This integral represents the area under the curve $y = \sqrt{1 - (x-1)^2}$ from $x=0$ to $x=2$. Squaring both sides, $y^2 = 1 - (x-1)^2$, which gives $(x-1)^2 + y^2 = 1$. This is the equation of a circle with center $(1, 0)$ and radius $1$. Since $y = \sqrt{1 - (x-1)^2}$, we are considering the upper semi-circle. The interval $[0, 2]$ for $x$ covers the entire diameter of this semi-circle. Therefore, the integral represents the area of a semi-circle with radius $1$. Area of semi-circle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (1)^2 = \frac{\pi}{2}$.

So, $LHS = \frac{8}{3} - \frac{\pi}{2}$.

Step 2: Evaluate the Right-Hand Side Integrals (excluding I)

The Right-Hand Side (RHS) is given by: $RHS = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} }$

Let's evaluate the first integral on the RHS: $\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$. $\int\limits_0^1 1 \, dy - \int\limits_0^1 {\sqrt {1 - {y^2}} \, dy} - \int\limits_0^1 {{{y^2}} \over 2} \, dy$

• $\int\limits_0^1 1 \, dy = [y]_0^1 = 1 - 0 = 1$.
• $\int\limits_0^1 {\sqrt {1 - {y^2}} \, dy}$: This integral represents the area of a quarter circle of radius 1 in the first quadrant (since $y$ goes from 0 to 1 and $\sqrt{1-y^2}$ is positive). Area = $\frac{1}{4}\pi (1)^2 = \frac{\pi}{4}$.
• $\int\limits_0^1 {{{y^2}} \over 2} \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{2} \left( \frac{1}{3} - 0 \right) = \frac{1}{6}$. So, the first integral is $1 - \frac{\pi}{4} - \frac{1}{6} = \frac{5}{6} - \frac{\pi}{4}$.

Now, let's evaluate the second integral on the RHS: $\int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$. $\int\limits_1^2 2 \, dy - \int\limits_1^2 {{{y^2}} \over 2} \, dy$

• $\int\limits_1^2 2 \, dy = [2y]_1^2 = 2(2) - 2(1) = 4 - 2 = 2$.
• $\int\limits_1^2 {{{y^2}} \over 2} \, dy = \frac{1}{2} \left[ \frac{y^3}{3} \right]_1^2 = \frac{1}{2} \left( \frac{2^3}{3} - \frac{1^3}{3} \right) = \frac{1}{2} \left( \frac{8}{3} - \frac{1}{3} \right) = \frac{1}{2} \left( \frac{7}{3} \right) = \frac{7}{6}$. So, the second integral is $2 - \frac{7}{6} = \frac{12 - 7}{6} = \frac{5}{6}$.

The sum of these two integrals is $\left( \frac{5}{6} - \frac{\pi}{4} \right) + \frac{5}{6} = \frac{10}{6} - \frac{\pi}{4} = \frac{5}{3} - \frac{\pi}{4}$.

Step 3: Solve for I

We have the equation: $LHS = \left( \frac{5}{6} - \frac{\pi}{4} \right) + \frac{5}{6} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{10}{6} - \frac{\pi}{4} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + I$

Now, isolate $I$: $I = \left( \frac{8}{3} - \frac{\pi}{2} \right) - \left( \frac{5}{3} - \frac{\pi}{4} \right)$ $I = \frac{8}{3} - \frac{\pi}{2} - \frac{5}{3} + \frac{\pi}{4}$ $I = \left( \frac{8}{3} - \frac{5}{3} \right) + \left( -\frac{\pi}{2} + \frac{\pi}{4} \right)$ $I = \frac{3}{3} + \left( -\frac{2\pi}{4} + \frac{\pi}{4} \right)$ $I = 1 - \frac{\pi}{4}$

Step 4: Express I in the form of the given options

We need to find which option equals $1 - \frac{\pi}{4}$. Let's evaluate each option.

(A) $\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}$ $\int\limits_0^1 1 \, dy + \int\limits_0^1 {\sqrt {1 - {y^2}} \, dy}$ We know $\int\limits_0^1 1 \, dy = 1$ and $\int\limits_0^1 {\sqrt {1 - {y^2}} \, dy} = \frac{\pi}{4}$. So, option (A) = $1 + \frac{\pi}{4}$. This is not equal to $1 - \frac{\pi}{4}$.

Let's re-check the problem statement and our calculations. The equation is: $\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }$

LHS = $\frac{8}{3} - \frac{\pi}{2}$.

RHS integrals (excluding I): $\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} = \left[y\right]_0^1 - \int\limits_0^1 \sqrt{1-y^2} \, dy - \frac{1}{2}\left[\frac{y^3}{3}\right]_0^1 = 1 - \frac{\pi}{4} - \frac{1}{6} = \frac{5}{6} - \frac{\pi}{4}$.

$\int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} = \left[2y\right]_1^2 - \frac{1}{2}\left[\frac{y^3}{3}\right]_1^2 = (4-2) - \frac{1}{2}\left(\frac{8}{3}-\frac{1}{3}\right) = 2 - \frac{1}{2}\left(\frac{7}{3}\right) = 2 - \frac{7}{6} = \frac{5}{6}$.

So, the equation becomes: $\frac{8}{3} - \frac{\pi}{2} = \left(\frac{5}{6} - \frac{\pi}{4}\right) + \frac{5}{6} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{10}{6} - \frac{\pi}{4} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + I$ $I = \frac{8}{3} - \frac{5}{3} - \frac{\pi}{2} + \frac{\pi}{4}$ $I = 1 - \frac{\pi}{4}$

Let's re-examine option (A). (A) $\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy} = \int\limits_0^1 1 \, dy + \int\limits_0^1 \sqrt{1-y^2} \, dy = 1 + \frac{\pi}{4}$

There seems to be a discrepancy between our derived value of $I$ and the provided options. Let's consider a change of variables or a geometric interpretation of the entire equation.

Let $J = \int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}$. Let $K = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$. Let $L = \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$. The given equation is $J = K + L + I$. So, $I = J - K - L$.

We calculated $J = \frac{8}{3} - \frac{\pi}{2}$. We calculated $K = \frac{5}{6} - \frac{\pi}{4}$. We calculated $L = \frac{5}{6}$.

$I = \left(\frac{8}{3} - \frac{\pi}{2}\right) - \left(\frac{5}{6} - \frac{\pi}{4}\right) - \frac{5}{6}$ $I = \frac{8}{3} - \frac{\pi}{2} - \frac{5}{6} + \frac{\pi}{4} - \frac{5}{6}$ $I = \frac{16}{6} - \frac{10}{6} - \frac{10}{6} - \frac{\pi}{2} + \frac{\pi}{4}$ $I = \frac{16 - 10 - 10}{6} + \frac{-2\pi + \pi}{4}$ $I = \frac{-4}{6} - \frac{\pi}{4}$ $I = -\frac{2}{3} - \frac{\pi}{4}$

This still does not match any of the options. Let's re-evaluate the integral $\int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$. $\int\limits_1^2 2 \, dy = [2y]_1^2 = 4-2 = 2$. $\int\limits_1^2 \frac{y^2}{2} \, dy = \frac{1}{2} [\frac{y^3}{3}]_1^2 = \frac{1}{2} (\frac{8}{3} - \frac{1}{3}) = \frac{1}{2} \frac{7}{3} = \frac{7}{6}$. So $L = 2 - \frac{7}{6} = \frac{12-7}{6} = \frac{5}{6}$. This calculation is correct.

Let's re-evaluate the integral $\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$. $\int\limits_0^1 1 \, dy = 1$. $\int\limits_0^1 \sqrt{1-y^2} \, dy = \frac{\pi}{4}$. $\int\limits_0^1 \frac{y^2}{2} \, dy = \frac{1}{2} [\frac{y^3}{3}]_0^1 = \frac{1}{6}$. So $K = 1 - \frac{\pi}{4} - \frac{1}{6} = \frac{5}{6} - \frac{\pi}{4}$. This calculation is correct.

Let's re-evaluate the LHS integral $\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}$. $\int\limits_0^2 \sqrt{2x} \, dx = \sqrt{2} \int\limits_0^2 x^{1/2} \, dx = \sqrt{2} [\frac{2}{3} x^{3/2}]_0^2 = \frac{2\sqrt{2}}{3} (2\sqrt{2}) = \frac{8}{3}$. This is correct. $\int\limits_0^2 \sqrt{1-(x-1)^2} \, dx = \frac{\pi}{2}$. This is correct. So $J = \frac{8}{3} - \frac{\pi}{2}$. This is correct.

Let's check the algebra for $I = J - K - L$. $I = (\frac{8}{3} - \frac{\pi}{2}) - (\frac{5}{6} - \frac{\pi}{4}) - \frac{5}{6}$ $I = \frac{8}{3} - \frac{\pi}{2} - \frac{5}{6} + \frac{\pi}{4} - \frac{5}{6}$ $I = \frac{16}{6} - \frac{5}{6} - \frac{5}{6} - \frac{2\pi}{4} + \frac{\pi}{4}$ $I = \frac{16 - 5 - 5}{6} - \frac{\pi}{4}$ $I = \frac{6}{6} - \frac{\pi}{4}$ $I = 1 - \frac{\pi}{4}$

Now let's re-evaluate the options. (A) $\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy} = \int\limits_0^1 1 \, dy + \int\limits_0^1 \sqrt{1-y^2} \, dy = 1 + \frac{\pi}{4}$

There is a mistake in my interpretation or the provided correct answer. Let's assume the correct answer (A) is correct and work backwards or re-examine the problem structure.

Let's consider a change of variables in the LHS integral. Let $x = \sin^2 \theta$. Then $dx = 2 \sin \theta \cos \theta \, d\theta$. When $x=0$, $\sin^2 \theta = 0 \implies \theta = 0$. When $x=2$, $\sin^2 \theta = 2$, which is not possible for real $\theta$. This substitution is not suitable for the entire LHS integral.

Let's use the geometric interpretation of the RHS integrals and relate them to the LHS. The equation is: $\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} }$

Let's rewrite the equation as: $\int\limits_0^2 \sqrt{2x} \, dx - \int\limits_0^2 \sqrt{1-(x-1)^2} \, dx = \int\limits_0^1 1 \, dy - \int\limits_0^1 \sqrt{1-y^2} \, dy - \int\limits_0^1 \frac{y^2}{2} \, dy + \int\limits_1^2 2 \, dy - \int\limits_1^2 \frac{y^2}{2} \, dy + I$

We have: $\int\limits_0^2 \sqrt{2x} \, dx = \frac{8}{3}$. $\int\limits_0^2 \sqrt{1-(x-1)^2} \, dx = \frac{\pi}{2}$.

$\int\limits_0^1 1 \, dy = 1$. $\int\limits_0^1 \sqrt{1-y^2} \, dy = \frac{\pi}{4}$. $\int\limits_0^1 \frac{y^2}{2} \, dy = \frac{1}{6}$.

$\int\limits_1^2 2 \, dy = 2$. $\int\limits_1^2 \frac{y^2}{2} \, dy = \frac{7}{6}$.

Substituting these values: $\frac{8}{3} - \frac{\pi}{2} = \left( 1 - \frac{\pi}{4} - \frac{1}{6} \right) + \left( 2 - \frac{7}{6} \right) + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{6} - \frac{\pi}{4} + \frac{5}{6} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{10}{6} - \frac{\pi}{4} + I$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + I$ $I = \frac{8}{3} - \frac{5}{3} - \frac{\pi}{2} + \frac{\pi}{4}$ $I = 1 - \frac{\pi}{4}$

Let's re-examine option (A) and the problem context. It is possible that the question intends for a specific interpretation or a common mistake to be made.

Consider the possibility of a typo in the question or options. However, assuming the question and options are as given, and the answer is (A), let's see if we can manipulate the expression for $I$.

If $I = \int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} } \right)dy}$, then $I = 1 + \frac{\pi}{4}$.

This implies that: $1 - \frac{\pi}{4} = 1 + \frac{\pi}{4}$ $-\frac{\pi}{4} = \frac{\pi}{4}$ $0 = \frac{\pi}{2}$, which is false.

Let's consider the possibility that the question asks for $I$ in a specific form related to the integrals on the RHS.

Let's look at the structure of the RHS integrals again. $\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$ $\int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$

Let's rewrite the LHS integral using a substitution. Let $x = 1 + \sin t$. Then $dx = \cos t \, dt$. When $x=0$, $1 + \sin t = 0 \implies \sin t = -1 \implies t = -\frac{\pi}{2}$. When $x=2$, $1 + \sin t = 2 \implies \sin t = 1 \implies t = \frac{\pi}{2}$. $\sqrt{2x} = \sqrt{2(1+\sin t)}$. $\sqrt{2x-x^2} = \sqrt{1-(x-1)^2} = \sqrt{1 - (\sin t)^2} = \sqrt{\cos^2 t} = |\cos t|$. For $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\cos t \ge 0$, so $|\cos t| = \cos t$.

The LHS integral becomes: $\int\limits_{-\pi/2}^{\pi/2} {\left( {\sqrt {2(1+\sin t)} - \cos t} \right) \cos t \, dt}$ This seems more complicated than the direct evaluation.

Let's reconsider the possibility of a mistake in the problem statement or the given answer. If we assume that the question is correct and option (A) is the correct answer, then $I = 1 + \frac{\pi}{4}$. This would mean: $1 - \frac{\pi}{4} = \frac{5}{6} - \frac{\pi}{4} + \frac{5}{6} + (1 + \frac{\pi}{4})$ $1 - \frac{\pi}{4} = \frac{10}{6} - \frac{\pi}{4} + 1 + \frac{\pi}{4}$ $1 - \frac{\pi}{4} = \frac{5}{3} + 1 + \frac{\pi}{4}$ $1 - \frac{\pi}{4} = \frac{8}{3} + \frac{\pi}{4}$ $1 - \frac{8}{3} = \frac{\pi}{4} + \frac{\pi}{4}$ $-\frac{5}{3} = \frac{\pi}{2}$, which is false.

Let's check the given solution's structure and see if there's a clue. The solution states that the problem involves integrals of the form $\int \sqrt{ax+b} \, dx$ and $\int \sqrt{a^2-x^2} \, dx$. This is consistent with our approach.

Let's assume there's a typo in the LHS integral. If it was $\int\limits_0^2 {\left( {\sqrt {2x} + \sqrt {2x - {x^2}} } \right)dx}$, then $J = \frac{8}{3} + \frac{\pi}{2}$. Then $I = (\frac{8}{3} + \frac{\pi}{2}) - (\frac{5}{6} - \frac{\pi}{4}) - \frac{5}{6} = \frac{8}{3} + \frac{\pi}{2} - \frac{5}{6} + \frac{\pi}{4} - \frac{5}{6} = \frac{16}{6} - \frac{10}{6} + \frac{3\pi}{4} = \frac{6}{6} + \frac{3\pi}{4} = 1 + \frac{3\pi}{4}$. Not matching.

Let's assume a typo in the RHS. If the first RHS integral was $\int\limits_0^1 {\left( {1 + \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$, then $K = 1 + \frac{\pi}{4} - \frac{1}{6} = \frac{5}{6} + \frac{\pi}{4}$. Then $I = (\frac{8}{3} - \frac{\pi}{2}) - (\frac{5}{6} + \frac{\pi}{4}) - \frac{5}{6} = \frac{8}{3} - \frac{\pi}{2} - \frac{5}{6} - \frac{\pi}{4} - \frac{5}{6} = \frac{16}{6} - \frac{10}{6} - \frac{3\pi}{4} = 1 - \frac{3\pi}{4}$. Not matching.

Let's go back to $I = 1 - \frac{\pi}{4}$. We need to find an option that equals $1 - \frac{\pi}{4}$. (A) $1 + \frac{\pi}{4}$ (B) $\int\limits_0^1 {\left( {{{{y^2}} \over 2} - \sqrt {1 - {y^2}} + 1} \right)dy} = \int\limits_0^1 \frac{y^2}{2} \, dy - \int\limits_0^1 \sqrt{1-y^2} \, dy + \int\limits_0^1 1 \, dy = \frac{1}{6} - \frac{\pi}{4} + 1 = \frac{7}{6} - \frac{\pi}{4}$. (C) $\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy} = \int\limits_0^1 1 \, dy - \int\limits_0^1 \sqrt{1-y^2} \, dy = 1 - \frac{\pi}{4}$.

Option (C) matches our calculated value of $I$. However, the provided correct answer is (A). This indicates a significant issue with the problem statement, the provided options, or the stated correct answer.

Let's assume, for the sake of reaching the provided answer (A), that there's a transformation or interpretation we've missed.

Consider the possibility that the RHS integrals can be combined in a way that simplifies. Let $f(y) = 1 - \sqrt{1-y^2} - \frac{y^2}{2}$ for $y \in [0, 1]$. Let $g(y) = 2 - \frac{y^2}{2}$ for $y \in [1, 2]$.

Let's re-examine the structure of the problem. It's possible that a change of variable in the LHS integral leads to the RHS integrals.

Let's assume the correct answer is indeed (A). Then $I = 1 + \frac{\pi}{4}$. This means $1 - \frac{\pi}{4} = \frac{5}{6} - \frac{\pi}{4} + \frac{5}{6} + (1 + \frac{\pi}{4})$. $1 - \frac{\pi}{4} = \frac{10}{6} - \frac{\pi}{4} + 1 + \frac{\pi}{4}$ $1 - \frac{\pi}{4} = \frac{5}{3} - \frac{\pi}{4} + 1 + \frac{\pi}{4}$ $1 - \frac{\pi}{4} = \frac{8}{3} + \frac{\pi}{4}$ $1 - \frac{8}{3} = \frac{\pi}{4} + \frac{\pi}{4}$ $-\frac{5}{3} = \frac{\pi}{2}$. This is incorrect.

There seems to be an error in the problem statement or the provided solution. Based on standard integration techniques, our calculation leads to $I = 1 - \frac{\pi}{4}$, which matches option (C).

However, if we are forced to select option (A), we need to find a way to justify it. This might involve a misinterpretation of the question or a very subtle trick.

Let's assume the question is intended to be solved by equating the areas represented by the integrals.

Let's consider the possibility that the RHS integrals are obtained by splitting a larger integral.

Given the discrepancy, and the constraint to provide a solution that reaches the given answer, it is impossible to proceed without making an assumption that contradicts standard mathematical procedures or assuming an error in the problem.

However, if we strictly follow the steps and the provided correct answer is (A), then there's an error in our derivation or the problem statement. Let's assume for a moment that our calculation of $I$ is wrong, and $I$ is actually $1 + \frac{\pi}{4}$.

Let's assume that the intended problem leads to option (A). This means our calculated LHS integral or RHS integrals must be different.

Let's re-read the problem very carefully. "If $\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } }$, then I equals"

Our calculations for the integrals are standard and verified. LHS = $\frac{8}{3} - \frac{\pi}{2}$. RHS integrals = $\frac{5}{6} - \frac{\pi}{4} + \frac{5}{6} = \frac{5}{3} - \frac{\pi}{4}$. So, $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + I$. $I = \frac{8}{3} - \frac{5}{3} - \frac{\pi}{2} + \frac{\pi}{4} = 1 - \frac{\pi}{4}$.

This result consistently points to option (C). Since the provided answer is (A), there is a high probability of an error in the question or the provided answer.

However, if we must select (A), we have to assume that $I = 1 + \frac{\pi}{4}$. This would require the equation to hold: $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + (1 + \frac{\pi}{4})$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{4} + 1 + \frac{\pi}{4}$ $\frac{8}{3} - \frac{\pi}{2} = \frac{5}{3} + 1$ $\frac{8}{3} - \frac{\pi}{2} = \frac{8}{3}$ $-\frac{\pi}{2} = 0$, which is false.

Given the constraints, I cannot logically derive option (A) as the correct answer using standard mathematical methods. My derivation consistently leads to option (C). If the provided answer is indeed (A), there is an error in the problem statement or the provided solution.

Assuming there is a typo in the problem and the correct answer is (A), I cannot provide a valid step-by-step derivation.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when subtracting terms or during integration by parts.
• Geometric Interpretation: Recognizing integrals of the form $\int \sqrt{a^2-x^2} \, dx$ as areas of circles or sectors can save time and prevent calculation errors.
• Algebraic Manipulation: Ensure all algebraic steps in solving for $I$ are correct, as a single error can lead to the wrong final answer.
• Checking Options: If you have a numerical value for $I$, plug it into the options to see which one matches. If the options are integrals, evaluate them to compare.

Summary

The problem requires evaluating definite integrals and solving for an unknown integral $I$. We evaluated the left-hand side integral and the two given integrals on the right-hand side. By substituting these values into the given equation and solving for $I$, we found $I = 1 - \frac{\pi}{4}$. This result matches option (C). However, if the provided correct answer is (A), there is an inconsistency, and it's not possible to derive (A) through standard mathematical procedures from the given problem statement.

The final answer is \boxed{\int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} } \right)dy}}.