Definite integration practice · Mixed difficulty

Option D is correct because by finding f(x) and g(x), we can evaluate the integral $\int_0^1 f(x) g(x) d x$ which results in $e-\frac{1}{2} e^2-\frac{3}{2}$.

The correct answer is D ($e^2+1$) because after using integration by parts and the given condition $f'(x)=f'(2-x)$, the integral evaluates to $e^2+1$.

Option C ($\frac{\pi}{8} \log 2$) is correct because it is the result of evaluating the definite integral using properties of logarithms and trigonometric functions after applying a substitution.

Option A ($\frac{1}{2}\,\tan \,1$) is correct because the limit can be expressed as a definite integral which evaluates to $\frac{1}{2} \tan 1$.

Option D (4) is correct because the function f(x) is increasing in the given interval, so the global maximum occurs at x=2, and f(2) = 4.

Option A ($\frac{{ - \,5\pi }}{3}$) is correct because the integral of the greatest integer function of 2sin(x) from pi to 2pi is calculated by splitting the integral into intervals where the greatest integer function has a constant value.

Option A ($\log \,\frac{4}{3}$) is correct because after substitution and partial fraction decomposition, the definite integral evaluates to this value.

Option B ($a$) is correct because the sum of the integrals can be combined into a single integral from 0 to 100, which is given as $a$.

Option D is correct because by using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ and simplifying the integral, we get the result $\frac{1}{n+1} - \frac{1}{n+2}$.

Option C $\pi$ is correct because the integral evaluates to $\pi$ after a suitable trigonometric substitution.

Option C ($\frac{1}{9} < {I^2} < \frac{1}{8}$) is correct because it accurately bounds the square of the integral based on the function's increasing nature within the given interval.

Option B ($\pi^2$) is correct because after simplifying the integral using properties of even and odd functions, the final result is $\pi^2$.

Option B (0) is correct because the integral can be simplified using the property $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab})$ and the substitution $x \rightarrow 1-x$.

Option D (1) is correct because L'Hopital's rule is applied to evaluate the limit.

Option C ($n / 2$) is correct because the integral of the fractional part of x from 0 to n is equal to n/2 due to the periodic nature of the function.

Option C is correct because integrating by parts and using the given information leads to the value -3.

Option C ($\frac{\pi}{12}$) is correct because after applying the property of definite integrals and simplifying, the integral evaluates to $\frac{\pi}{12}$.

Option A ($4\pi$) is correct. The integral simplifies due to the odd function and trigonometric identities, resulting in the value $4\pi$.

Option D ($- \pi$) is correct because the integral simplifies to integrating -1 over the given interval due to the properties of the greatest integer function and sine function.

Option A ($\frac{\pi}{4}$) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to $\frac{\pi}{4}$.

Option B (2) is correct because the integral is evaluated by splitting it into two parts based on the absolute value function and then integrating each part.

Option A (12 \x² F ,( \x² ) - ∫ F ,( \x² ) dx² $\frac{1}{2}\\ {\x^2}\\ F\\,\left( {\x^2} \right)\\ - \\ \int F\\,\left( {\x^2} \right)\\ d{x^2}\\}\;$) is correct because it is derived using integration by parts with appropriate substitution.

Option A (0) is correct because the integral of the product of cosine functions with different integer multiples over a full period is zero.

Option D (None of these) is correct because the integral evaluates to $\sin(1) - \frac{1}{4}\sin(2)$, which is not among the given options.

Option C (0) is correct because the function x|x| is an odd function, and the integral of an odd function over a symmetric interval is zero.

Option D ($\frac{16}{3} \pi-2 \sqrt{3}$) is correct. The integral is solved using integration by parts and a suitable substitution to simplify the expression.

Option D ($\left[ { - \,\,\sqrt 2 - \,\,\sqrt 3 + 5} \right]$) is correct because the integral of the greatest integer function of x squared over the interval [0, 2] evaluates to this expression.

Option B (1) is correct because the integral simplifies to 1 after using trigonometric identities and evaluating the definite integral.

Option C (-5) is correct because it is the result of using the properties of definite integrals to find the value of the required integral.

Option D, $6\;\mathrm{\pi \;sq\;unit}$, is the correct area of the ellipse because the area is calculated using the formula $\pi ab$, where a and b are the semi-major and semi-minor axes.