Definite integration practice · Mixed difficulty

Option B ($\frac{\mathrm{\pi}}{6}$) is correct because the integral can be solved by direct substitution and evaluating the limits.

Option C ($3I$) is correct because the integral $\int_3^{3+3T} f(2x) dx$ simplifies to $3I$ using the periodicity of the function $f(x)$.

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

Option D ($k$) is correct because the integral of odd powers of x from -3 to 3 is zero, leaving only the integral of the constant k.

The correct answer is B (${\log _e}\,16$) because it is the result of simplifying the integral and evaluating it within the given limits.

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 B (0) is correct because applying L'Hopital's rule to the limit results in an expression that simplifies to 0 as x approaches 0.

Option D (28/3) is correct because it involves splitting the integral into intervals where $|1-x^2|$ is either $(1-x^2)$ or $(x^2-1)$, then integrating and summing the results.

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 B is correct because the integral is evaluated by splitting it into two parts based on the sign of the logarithm function and then using a simple substitution.

Option C ($\frac{1}{6}$) is correct because the integral can be simplified by substitution, leading to an arctangent integral which evaluates to the given value.

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 $\frac{1}{{{2^n}}}$.

Option D (I2 < I1) is correct because for x in [0, 1], 2^(x3) < 2^(x2), implying that the integral I2 is less than I1.

Option C ($\mathrm{19}$) is correct. The integral is evaluated by summing the integrals over the intervals [n, n+1] from -2 to 3.

Option D ($\frac{1}{7}$) is correct because the sum of the integrals simplifies to a single integral that can be easily evaluated.

Option B ($\frac{{2\pi}}{3}$) is correct because the integral can be solved using the property of definite integrals and trigonometric identities to simplify the expression.

Option D (1) is correct because it is found by evaluating the integral of the difference between g(x) and f(x) over the given interval.

Option D (8 + $\frac{3 \sqrt{3}}{2} \pi$) is correct because after applying the appropriate substitutions and evaluating, the integral results in 8 + $\frac{3 \sqrt{3}}{2} \pi$.

Option B is correct. The integral is solved using substitution and simplifying the expression, resulting in 4 - $\pi$.

Option B ($\frac{\pi}{2}$) is correct because using the property $\int_{-a}^a f(x) dx = \int_{-a}^a f(-x) dx$ simplifies the integral.

The correct answer is C ($\frac{1}{{n + 1}}\\ - \frac{1}{{n + 2}}$) because it is the result of evaluating the definite integral using substitution or Beta function properties.

Option B is correct because the integral can be solved using the property $\int_a^b \frac{f(x)}{f(x)+f(a+b-x)} dx = \frac{b-a}{2}$.

Option C ($\mathrm{1}\,\mathrm{/}\,\mathrm{4}$) is correct because the integral evaluates to $\frac{1}{4}\,\log \left( {\frac{{3 + 2\,\sqrt 3 }}{3}} \right)$, thus $k = \frac{1}{4}$.

Option B (2) is correct because it is found by using integration by parts twice and simplifying the resulting expression.

Option A is correct because integrating by parts simplifies the integral, leading to the result $2 \tan \frac{3 \pi}{8}$.

Option C ($75\pi$) is correct because it is the result of evaluating the definite integral using properties of definite integrals.

Option B ($\frac{\pi}{2} \int\limits_{0}^\pi f(\sin x) d x$) is correct because it applies the property of definite integrals where $\int_0^a x f(sin(x))dx = \frac{a}{2}\int_0^a f(sin(x))dx$.

Option D ($\frac{1}{{{b_3} - {b_2}}},\frac{1}{{{b_4} - {b_3}}},\frac{1}{{{b_5} - {b_4}}}$ are in an A.P. with common difference –2) is correct because by evaluating the integral and finding the differences between consecutive terms, we can show that the reciprocals form an arithmetic progression.

Option B (2) is correct because the integral simplifies to the integral of |sin x| over the interval [-$\pi/2$, $\pi/2$], which evaluates to 2.