Definite integration practice · Medium

The correct answer is D (None of these) because the value of the integral is $\frac{116}{15}$, which is not among the options.

Option A (1) is correct because the integral of the fractional part of x from 0 to 2 is equal to 1.

Option B ($\frac{\pi }{2}\\ - \\,\log \\,2$) is correct because the integral is solved by using the property $\cot^{-1}(x) = \tan^{-1}(1/x)$ and simplifying the integrand using the tangent addition formula.

Option C (√2 - 1) is correct because the integral of the greatest integer function [x²] from 0 to √2 can be split into two integrals, one from 0 to 1 where [x²] = 0 and another from 1 to √2 where [x²] = 1.

Option B ($\frac{{1}}{{6}} n(n-1)(4 n+1)$) is correct because it represents the sum of the greatest integer of the square root of x over the given interval.

Option B ($\pm \,2$) is correct because the common difference of the A.P. is found by evaluating the integral and solving the resulting equation.

Option A ($\pm 2\\,\sqrt {\frac{{\sin \\,\alpha }}{\alpha }}$) is correct because after simplifying the equation using properties of odd functions and evaluating the integral, the value of x can be found.

Option B ($-\log \left|1+\sqrt{{2-x^2}}\right|+c$) is correct because the integral can be solved by substituting $t = \sqrt{{2-x^2}}$.

Option B (\$\frac{{\pi}}{{8}}-\frac{{\ln 2}}{{4}}$\$) is correct. By using substitution and integration by parts, the definite integral is evaluated to be$$\frac{{π}}{{8}}$-$\frac{{\ln 2}}{{4}}$.

Option B (I < 2/3 and J < 2) is correct because it follows from comparing the integrands with simpler functions and evaluating the resulting integrals.

Option A (3) is correct because the integral of the fractional part of the square root of x from 0 to 100 can be calculated by subtracting the sum of integrals from the original integral, resulting in the value 3.

Option D ($\frac{1}{I_2+I_4}, \frac{1}{I_3+I_5}, \frac{1}{I_4+I_6}$ are in A.P) is correct because by using integration by parts, we can find a recurrence relation for $I_n$ and show that the reciprocals are in arithmetic progression.

Option C is correct. The area enclosed by the curves is found by integrating the difference between the two functions from -2 to 2, which results in 2π - 4/3.

Option B (37/6) is correct because it is the total area calculated by splitting the integral into two parts and summing them up.

Option A ($\frac{2n}{n^2-m^2}$) is correct because the integral evaluates to $\frac{2n}{n^2-m^2}$ when $(n-m)$ is odd and $|m| \neq |n|$.

Option B (-1+√5) is correct because the limit of the Riemann sum is converted to a definite integral and evaluated correctly.

Option B (2) is correct because by breaking the integral into intervals where the greatest integer function is constant, the integral evaluates to 2.

Option A ($\frac{1}{{n\, - \,1}}$) is correct because by using integration by parts and simplifying, the expression ${u_n}\, + \,{u_{n - 2}}$ simplifies to $\frac{1}{{n\, - \,1}}$.

Option B is correct because the integral can be solved by breaking it into simpler integrals and using appropriate substitutions.

Option C (-$\frac{{3}}{{4}}$) is correct because it is the value of 'a' that minimizes the area under the given curve between the specified limits, found by completing the square and identifying the vertex.

Option C ($\ell n 11$) is correct because after substituting $y = e^{f(2t)}$ and solving the integral, we find that $f(4) = \ell n 11$.

Option B is correct because the integral of a periodic function over n periods is n times the integral over one period.

Option A ($1-\frac{\pi}{4}$) is correct because the integral is solved by simplifying the expression and using trigonometric identities.

Option A ($\frac{{1}}{{\mathrm{{em}}}}$) is correct because it is the result of evaluating the limit using Stirling's approximation or by converting the limit into an integral.

Option D (√2 g(1)=g(0)) is correct because it is derived from the properties of the integral and the even/odd nature of the functions involved, leading to the relationship between g(1) and g(0).

Option D ($\frac{{\pi}}{{4}}$) is correct. The integral is solved using the property of definite integrals where $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.

Option B ($\frac{3 \pi}{8}$) is correct because after using trigonometric identities and integrating, the result is $\frac{3 \pi}{8}$.

Option D ($4-\pi$) is correct because the integral is solved using substitution and trigonometric inverse functions.

Option D (-5) is correct because the integral of an odd function from -1 to 0 is the negative of the integral from 0 to 1, given that $\int_0^1 f(x) dx = 5$.

Option C ($\frac{1}{{{2^n}}}$) is correct because after applying trigonometric identities and properties of definite integrals, the value of $\lambda$ is found to be $\frac{1}{{{2^n}}}$.