Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Completing the square in the denominator yields $(x-1)^2 + 3$, which suggests the trigonometric substitution $x-1 = \sqrt{3} \tan \theta$, where $dx = \sqrt{3} \sec^2 \theta \, d\theta$. Changing the integration limits from $x=1$ to $x=2$ corresponds to $\theta=0$ to $\theta=\pi/6$. The integral becomes:

$\int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta \, d\theta}{(3 \sec^2 \theta)^{3/2}} = \int_{0}^{\pi/6} \frac{\sqrt{3} \sec^2 \theta \, d\theta}{3\sqrt{3} \sec^3 \theta} = \frac{1}{3} \int_{0}^{\pi/6} \cos \theta \, d\theta$

Evaluating this expression results in $\frac{1}{3} [\sin \theta]_0^{\pi/6} = \frac{1}{3} (\frac{1}{2} - 0) = \frac{1}{6}$. Equating this result to the given expression $\frac{k}{k+5}$ yields $\frac{k}{k+5} = \frac{1}{6}$, which implies $6k = k+5$, so $5k = 5$, and therefore $k=1$.

The evaluation of limits of sums of powers as $n \to \infty$ relies on the integral approximation $\sum_{k=1}^n k^a \approx \int_0^n x^a dx = \frac{n^{a+1}}{a+1}$. Applying this approximation to the numerator, the sum $\sum_{k=1}^n k^a$ behaves asymptotically as $\frac{n^{a+1}}{a+1}$ for large values of $n$.

For the denominator, $(n+1)^{a-1} \approx n^{a-1}$ as $n \to \infty$, and the sum of the arithmetic series is given by $\sum_{k=1}^n (na+k) = n(na) + \frac{n(n+1)}{2} = n^2a + \frac{n^{2+}n}{2} = n^2(\frac{2a+1}{2}) + \frac{n}{2}$. As $n$ approaches infinity, the dominant term is $n^2(\frac{2a+1}{2})$, making the entire denominator expression asymptotically equivalent to $n^{a-1} \cdot n^2(\frac{2a+1}{2}) = n^{a+1}(\frac{2a+1}{2})$.

Substituting these asymptotic forms into the original limit expression yields the equation $\frac{n^{a+1}/(a+1)}{n^{a+1}(2a+1)/2} = \frac{2}{(a+1)(2a+1)}$. Setting this result equal to $\frac{1}{60}$ leads to $(a+1)(2a+1) = 120$, which simplifies to the quadratic equation $2a^2 + 3a - 119 = 0$. Factoring this quadratic as $(2a+17)(a-7) = 0$ provides the positive root $a = 7$.

The fundamental identity linking tangent and cotangent, defined as $\tan x + \cot x$, can be rewritten as $\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}$, which simplifies to $\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}$ and ultimately $\frac{1}{\sin x \cos x}$. Utilizing the double angle identity, $\sin 2x = 2 \sin x \cos x$, this expression becomes $\frac{2}{\sin 2x}$. Cubing this result produces $(\tan x + \cot x)^3 = \frac{8}{\sin^3 2x}$, which allows the integrand to be simplified by substituting the denominator, resulting in $\int_{\pi/12}^{\pi/4} \frac{8 \cos 2x}{8 / \sin^3 2x} \, dx$, or simply $\int_{\pi/12}^{\pi/4} \sin^3 2x \cos 2x \, dx$.

Solving this integral relies on substitution, letting $u = \sin 2x$, which makes the differential $du = 2 \cos 2x \, dx$, or $\cos 2x \, dx = \frac{1}{2} du$. Changing the limits of integration from $x$ values of $\pi/12$ and $\pi/4$ to $u$ values yields $\sin(\pi/6) = 1/2$ and $\sin(\pi/2) = 1$, respectively. Substituting these into the integral gives $\int_{1/2}^{1} \frac{u^3}{2} \, du$, which calculates to \frac{1}{2} \[\frac{u^4}{4}]_{1/2}^{1} = \frac{1}{8} (1 - \frac{1}{16})$, arriving at the final value of$\frac{15}{128}$.

Applying the integral property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ to the interval from $4$ to $10$ allows us to substitute $x$ with $14-x$. The denominator's quadratic expression $x^{2-}28x+196$ is equivalent to $(x-14)^2$, which transforms into $(-x)^2 = x^2$ after this substitution, while the original $x^2$ term in the numerator becomes $(14-x)^2$. Summing the original integral and its transformed version results in the integral of \frac{[x^2]\+$[(14-x)^2]$}{[(x-14)^2]$+$[x^2]$}$, which simplifies to$\int_{4}^{10} 1 dx = 6$. Since this sum equals$2I$, we find$I = 3$.

Differentiating the integral equation with respect to $x$ using the product rule and the Fundamental Theorem of Calculus simplifies the expression to the relationship $\int_{1}^{x} y(t) \, dt = x^2 y(x) - \int_{1}^{x} t y(t) \, dt$. A second differentiation with respect to $x$ removes the integral terms, yielding $y(x) = 2x y(x) + x^2 y'(x) - x y(x)$, which simplifies further to $y(x) = x y(x) + x^2 y'(x)$.

Rearranging this into the form $y(x)(1 - x) = x^2 \frac{dy}{dx}$ facilitates the separation of variables. Dividing by $y(x)x^2$ leads to $\frac{dy}{y} = (\frac{1}{x^2} - \frac{1}{x}) \, dx$. Integrating both sides results in $\ln|y| = -\frac{1}{x} - \ln|x| + \ln|C|$, which combines to $\ln|xy| = -\frac{1}{x} + \ln|C|$. Exponentiating both sides confirms that the function is $y(x) = \frac{C}{x} e^{-1/x}$.

The fundamental relationship between inverse trigonometric functions dictates that $\tan^{-1} u + \cot^{-1} u = \frac{\pi}{2}$ for any real number $u$. This identity allows the integral $\int_0^1 \tan^{-1}(1-x+x^2) dx$ to be rewritten by substituting the integrand with $\frac{\pi}{2} - \cot^{-1}(1-x+x^2)$, which separates the expression into $\int_0^1 \frac{\pi}{2} dx - \int_0^1 \cot^{-1}(1-x+x^2) dx$. Evaluating the constant integral results in $\frac{\pi}{2}$, and the second term is directly provided as $2\int_0^1 \tan^{-1} x dx$ based on the given equation.

Substituting this into the derived expression transforms the problem into calculating $\frac{\pi}{2} - 2\int_0^1 \tan^{-1} x dx$. By applying integration by parts to $\int_0^1 \tan^{-1} x dx$, where the derivative of $\tan^{-1} x$ is $\frac{1}{1+x^2}$ and the integral of $1$ is $x$, the result is $[x \tan^{-1} x]_0^1 - \int_0^1 \frac{x}{1+x^2} dx$. Evaluating these bounds yields $\frac{\pi}{4} - \frac{1}{2} \ln 2$. Multiplying this value by two and subtracting it from $\frac{\pi}{2}$ gives $\frac{\pi}{2} - 2(\frac{\pi}{4} - \frac{1}{2} \ln 2)$, which simplifies to $\log 2$.

To evaluate a limit involving a product of factors raised to a power of $1/n$, applying the natural logarithm transforms the product into a sum, which is the ideal form for identifying a Riemann integral. Let the given limit be $y$. Taking the logarithm converts the expression into the limit $\lim_{n\to \infty } \frac{1}{n} \sum_{r=1}^{2n} \ln\left(1 + \frac{r}{n}\right)$, which corresponds to the definite integral $\int_0^2 \ln(1 + x) \, dx$. Evaluating this integral via the substitution $t = 1 + x$ gives $\int_1^3 \ln t \, dt$, which solves to $[t \ln t - t]_1^3 = (3 \ln 3 - 3) - (0 - 1) = 3 \ln 3 - 2$. Consequently, the original limit $y$ is $e^{3 \ln 3 - 2}$, which simplifies to $\frac{27}{e^2}$.

This problem relies on the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. If we define $I$ as the integral $\int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(36-12x+x^2)} \, dx$, we can observe that the expression $36-12x+x^2$ is the square of $(6-x)$. Applying the property with $a+b = 2+4 = 6$ transforms $x$ into $6-x$, which effectively swaps the terms in the denominator, resulting in $I = \int_{2}^{4} \frac{\log(6-x)^2}{\log(6-x)^2 + \log x^2} \, dx$. Adding the two forms of the integral yields $2I = \int_{2}^{4} \frac{\log x^2 + \log(6-x)^2}{\log x^2 + \log(6-x)^2} \, dx$, which simplifies to $\int_{2}^{4} 1 \, dx$. Evaluating this gives $2I = [x]_{2}^{4} = 2$, and dividing by $2$ shows the value of the original integral is $1$.

The core principle here is recognizing that the expression under the square root, $1-4\sin(x/2)+4\sin^2(x/2)$, is a perfect square trinomial, which simplifies to $(1-2\sin(x/2))^2$. Since $\sqrt{u^2}$ is equivalent to $|u|$, the integral becomes the accumulation of $|1-2\sin(x/2)|$ across the interval $[0, \pi]$. Because the term inside the absolute value changes sign when $1-2\sin(x/2) = 0$, which occurs at $x/2 = \pi/6$ or $x = \pi/3$, the domain must be split into two parts to handle the absolute value correctly.

On the interval $[0, \pi/3]$, the expression $1-2\sin(x/2)$ is non-negative, so the integrand remains as is, leading to the integral $\int_{0}^{\pi/3} (1-2\sin(x/2)) \, dx$. Evaluating this yields $[x+4\cos(x/2)]_0^{\pi/3}$, which simplifies to $(\pi/3 + 2\sqrt{3}) - (0 + 4)$, or $\pi/3 + 2\sqrt{3} - 4$. In the subsequent interval $[\pi/3, \pi]$, $1-2\sin(x/2)$ is negative, so the integrand must be negated to $2\sin(x/2)-1$ to maintain a positive area. Evaluating $\int_{\pi/3}^{\pi} (2\sin(x/2)-1) \, dx$ gives $[-4\cos(x/2)-x]_{\pi/3}^{\pi}$, which resolves to $(-4\cos(\pi/2) - \pi) - (-4\cos(\pi/6) - \pi/3)$, or simply $-\pi + 2\sqrt{3} + \pi/3$, resulting in $2\sqrt{3} - 2\pi/3$. Combining these two evaluated segments, $(\pi/3 + 2\sqrt{3} - 4) + (2\sqrt{3} - 2\pi/3)$, results in the final value $4\sqrt{3} - 4 - \pi/3$.

The principle presented in the second statement is the standard King's property for definite integrals, which is entirely correct. To determine the value of the integral $I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$, we can apply this property by replacing $x$ with the sum of the limits minus $x$, which is $\pi/2 - x$. This transformation changes the integrand to $\frac{1}{1+\sqrt{\cot x}}$, which simplifies algebraically to $\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}$. Adding this transformed integral to the original integral yields:

$\int_{\pi/6}^{\pi/3} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} dx = \int_{\pi/6}^{\pi/3} 1 dx$

This simplified expression integrates to $\pi/3 - \pi/6$, resulting in a value of $\pi/6$. Since this calculation represents the sum of the original integral and its transformed version, it follows that $2I = \pi/6$, meaning the actual value of the integral $I$ is $\pi/12$. Because the first statement claims the integral is equal to $\pi/6$ instead of $\pi/12$, that statement is false.

The integral function $g(x+\pi) = \int_0^{x+\pi} \cos(4t) \, dt$ can be separated into two parts by splitting the interval of integration at $\pi$, which gives $\int_0^\pi \cos(4t) \, dt + \int_\pi^{x+\pi} \cos(4t) \, dt$. The first part is equal to $g(\pi)$, while the second integral can be simplified by substituting $t = u + \pi$, resulting in $\int_0^x \cos(4(u + \pi)) \, du$. Since $\cos(4u + 4\pi) = \cos(4u)$, this integral simplifies to $\int_0^x \cos(4u) \, du$, which is defined as $g(x)$. Combining these components yields $g(\pi) + g(x)$, and because $g(\pi) = \int_0^\pi \cos(4t) \, dt = [\frac{\sin(4t)}{4}]_0^\pi = 0$, this expression is mathematically equivalent to $g(x) - g(\pi)$.

Evaluating this definite integral is most straightforward when utilizing a trigonometric substitution and the definite integral property $\int_{0}^{a} f(\theta) \, d\theta = \int_{0}^{a} f(a - \theta) \, d\theta$. By setting $x = \tan \theta$, the differential $dx$ becomes $\sec^2 \theta \, d\theta$, and the interval from 0 to 1 transforms into the interval from 0 to $\pi/4$. Substituting these into the integral, the term $1+x^2$ becomes $1+\tan^2 \theta$, which simplifies to $\sec^2 \theta$. Since this factor cancels perfectly with the differential, the original expression reduces to the simpler form $I = \int_{0}^{\pi/4} 8 \log(1 + \tan \theta) \, d\theta$.

Applying the substitution $\theta = \pi/4 - \theta$ within this integral relies on the trigonometric identity for the tangent of a difference, where $\tan(\pi/4 - \theta)$ expands to $\frac{1 - \tan \theta}{1 + \tan \theta}$. Adding 1 to this value yields $\frac{2}{1 + \tan \theta}$, which allows the integrand to be rewritten as $8 \log \left( \frac{2}{1 + \tan \theta} \right)$. Using logarithm properties, this splits into $8 \log 2 - 8 \log(1 + \tan \theta)$.

The integral now consists of two parts: the integral of the constant $8 \log 2$ over the range from 0 to $\pi/4$, minus the original integral $I$. Integrating the constant over this range yields $8 \log 2$ multiplied by $\pi/4$, which simplifies to $2 \pi \log 2$. Therefore, the equation becomes $I = 2 \pi \log 2 - I$, establishing that $2I = 2 \pi \log 2$ and leaving the final result as $\pi \log 2$.

Integrating both sides of the condition $p'(x)=p'(1-x)$ with respect to $x$ leads to the relationship
$p(x) = -p(1-x) + C$
where the negative sign appears due to the chain rule applied to the inner function $1-x$. Substituting the boundary values $p(0)=1$ and $p(1)=41$ at $x=0$ yields $1 = -p(1) + C$, which simplifies to $1 = -41 + C$, fixing the constant at $C=42$. This establishes the symmetrical identity
$p(x) + p(1-x) = 42$
for any value of $x$ in the interval $[0, 1]$.

To determine the integral $I = \int_{0}^{1} p(x) dx$, we utilize the property that the definite integral of a function over an interval remains unchanged if the variable is replaced by the sum of the bounds minus the variable. We can express $2I$ as the sum of the integral of $p(x)$ and the integral of $p(1-x)$, which combines into a single integral:
$2I = \int_{0}^{1} (p(x) + p(1-x)) dx$
Substituting the derived identity into this expression results in $\int_{0}^{1} 42 dx$, which evaluates to $42$. Dividing by $2$ confirms that the value of the integral is $21$.

The definite integral property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ allows us to equate the integral $\int_{0}^{\pi} [\cot x] dx$ to $\int_{0}^{\pi} [-\cot x] dx$. Adding these two expressions yields $2\int_{0}^{\pi} [\cot x] dx = \int_{0}^{\pi} ([\cot x] + [-\cot x]) dx$. Since the greatest integer function identity $[y] + [-y] = -1$ holds for all non-integer values of $y$, the integrand simplifies to $-1$ across the interval $(0, \pi)$. Integrating $-1$ with respect to $x$ results in $-\pi$, which leads to the final value:
$I = -\frac{\pi}{2}$

The indefinite integral $\int \frac{dt}{t \sqrt{t^{2-}1}}$ evaluates directly to $\sec^{-1} t$. Applying this to the definite integral across the interval from $\sqrt{2}$ to $x$ yields $\sec^{-1} x - \sec^{-1} \sqrt{2} = \frac{\pi}{2}$. Given that $\sec^{-1} \sqrt{2}$ is $\frac{\pi}{4}$, the equation simplifies to $\sec^{-1} x = \frac{\pi}{2} + \frac{\pi}{4}$, which gives $\sec^{-1} x = \frac{3\pi}{4}$. Solving for $x$ results in $x = \sec \left( \frac{3\pi}{4} \right) = -\sqrt{2}$, a value which does not match any of the provided options.

The definition of $F(e)$ as $f(e) + f(1/e)$ utilizes the property that the integral $f(x) = \int_1^x \frac{\log t}{1+t} dt$ can be combined with its counterpart after applying a substitution. Replacing $t$ with $1/u$ in the second term, $f(1/e) = \int_1^{1/e} \frac{\log t}{1+t} dt$, involves the differential $dt = -1/u^2 du$ and changes the integration boundaries from $1$ and $1/e$ to $1$ and $e$, simplifying the integral to $\int_1^e \frac{\log u}{u(1+u)} du$. Adding this result to the first term $f(e) = \int_1^e \frac{\log t}{1+t} dt$ permits the integrands to combine into the expression $\frac{\log t}{1+t} + \frac{\log t}{t(1+t)}$, which simplifies algebraically to $\frac{\log t(t+1)}{t(1+t)} = \frac{\log t}{t}$. This leaves the integral $\int_1^e \frac{\log t}{t} dt$, and substituting $v = \log t$ transforms the expression into $\int_0^1 v dv$, which evaluates to $\frac{1}{2}$.

Bounding definite integrals by comparing their integrands to simpler, integrable functions is a reliable method for establishing these inequalities. For the integral $I$, since $\sin x < x$ for $x$ between 0 and 1, we obtain the inequality $\frac{\sin x}{\sqrt{x}} < \sqrt{x}$, which implies $I < \int_{0}^{1} \sqrt{x} \, dx = \frac{2}{3}$. Regarding $J$, noting that $\cos x < 1$ on the same interval leads to $\frac{\cos x}{\sqrt{x}} < \frac{1}{\sqrt{x}}$, which demonstrates that $J < \int_{0}^{1} x^{-1/2} \, dx = 2$. These results confirm that both inequalities $I < \frac{2}{3}$ and $J < 2$ must hold true.

The most efficient way to solve this integral is by utilizing the symmetry properties of even and odd functions, which allows the terms to cancel out or simplify significantly. Begin by substituting $t = x + \pi$, which shifts the limits of integration to $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and simplifies the term $\cos^2(x+3\pi)$ to $\cos^2(t+2\pi)$. Because $\cos^2(t+2\pi) = \cos^2 t$, the integral becomes $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (t^3 + \cos^2 t) \, dt$.

Since $t^3$ is an odd function integrated over a symmetric interval, its contribution to the integral is zero. The remaining expression, $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2 t \, dt$, involves an even function, which can be evaluated as $2 \int_{0}^{\frac{\pi}{2}} \cos^2 t \, dt$. Applying the trigonometric identity $\cos^2 t = \frac{1 + \cos 2t}{2}$ transforms the integral into $\int_{0}^{\frac{\pi}{2}} (1 + \cos 2t) \, dt$. Integrating this yields $t + \frac{\sin 2t}{2}$ evaluated from $0$ to $\frac{\pi}{2}$, which results in a final value of $\frac{\pi}{2}$.

The King's property of definite integrals, which states that $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$, is the most effective approach for evaluating this integral. By substituting $x$ with $(\pi - x)$, the original integral $I = \int_{0}^{\pi} x f(\sin x) dx$ transforms into $I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) dx$. Since $\sin(\pi - x) = \sin x$, the expression simplifies to $I = \int_{0}^{\pi} (\pi - x) f(\sin x) dx$, which can be expanded into $\pi \int_{0}^{\pi} f(\sin x) dx - I$. Adding $I$ to both sides results in $2I = \pi \int_{0}^{\pi} f(\sin x) dx$, yielding $I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) dx$.

Because the function $f(\sin x)$ is symmetric about $\frac{\pi}{2}$ on the interval $[0, \pi]$, the property $\int_{0}^{2a} g(x) dx = 2 \int_{0}^{a} g(x) dx$ allows us to rewrite $\int_{0}^{\pi} f(\sin x) dx$ as $2 \int_{0}^{\pi/2} f(\sin x) dx$. Substituting this result back into the previous equation gives $I = \frac{\pi}{2} \cdot 2 \int_{0}^{\pi/2} f(\sin x) dx$, which simplifies to $I = \pi \int_{0}^{\pi/2} f(\sin x) dx$. Finally, applying the identity $\sin(\frac{\pi}{2} - x) = \cos x$ within the definite integral confirms that $\int_{0}^{\pi/2} f(\sin x) dx$ is equivalent to $\int_{0}^{\pi/2} f(\cos x) dx$, arriving at the final value of $\pi \int_{0}^{\pi/2} f(\cos x) dx$.

The floor function $\lfloor x \rfloor$ remains constant between consecutive integers, meaning the integrand effectively shifts values at every integer point. To compute the integral $\int_{1}^{a} \lfloor x \rfloor f'(x) \, dx$, we decompose the range $[1, a]$ into sub-intervals $[n, n+1]$ for $n=1, 2, \dots, \lfloor a \rfloor - 1$, and one final interval $[ \lfloor a \rfloor, a]$. Integrating over these pieces allows us to express the total value as:
$\int_{1}^{2} 1 f'(x) \, dx + \int_{2}^{3} 2 f'(x) \, dx + \dots + \int_{\lfloor a \rfloor - 1}^{\lfloor a \rfloor} (\lfloor a \rfloor - 1) f'(x) \, dx + \int_{\lfloor a \rfloor}^{a} \lfloor a \rfloor f'(x) \, dx$
Evaluating these individual integrals using the fundamental theorem of calculus yields $1(f(2)-f(1)) + 2(f(3)-f(2)) + \dots + (\lfloor a \rfloor - 1)(f(\lfloor a \rfloor)-f(\lfloor a \rfloor - 1)) + \lfloor a \rfloor (f(a)-f(\lfloor a \rfloor))$. When we distribute these coefficients, the terms associated with each $f(n)$ for $n$ from $1$ to $\lfloor a \rfloor$ are combined. Specifically, the coefficient for $f(2)$ becomes $1-2 = -1$, and for any intermediate $f(n)$, the coefficient is $(n-1)-n = -1$. This simplifies the entire expression to $-f(1) - f(2) - f(3) - \dots - f(\lfloor a \rfloor) + \lfloor a \rfloor f(a)$, which is identical to $\lfloor a \rfloor f(a) - \{f(1)+f(2)+\dots + f(\lfloor a \rfloor)\}$.