Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Definite integrals involving the greatest integer function are best evaluated by partitioning the interval of integration at every integer boundary, as the integrand remains constant between these points. For the integral $\int_{0}^{3} \frac{e^{x}+e^{-x}}{[x]!} dx$, the integer boundaries are 1 and 2, which allow splitting the integral into three sub-intervals: $[0, 1)$, $[1, 2)$, and $[2, 3)$.

In the first interval $0 \le x < 1$, the value of $[x]$ is 0, so $[x]!$ becomes $0! = 1$. The integral simplifies to $\int_{0}^{1} (e^x + e^{-x}) dx$, which evaluates to $[e^x - e^{-x}]_0^1 = e - \frac{1}{e}$. For the second interval $1 \le x < 2$, the value of $[x]$ is 1, making $[x]!$ equal to $1! = 1$. The integral becomes $\int_{1}^{2} (e^x + e^{-x}) dx$, resulting in $[e^x - e^{-x}]_1^2 = (e^2 - \frac{1}{e^2}) - (e - \frac{1}{e})$. Finally, in the third interval $2 \le x < 3$, the value of $[x]$ is 2, so $[x]!$ becomes $2! = 2$. This leads to $\int_{2}^{3} \frac{e^x + e^{-x}}{2} dx$, which evaluates to \frac{1}{2} \[e^x - e^{-x}]_2^3 = \frac{1}{2}(e^3 - \frac{1}{e^3}) - \frac{1}{2}(e^2 - \frac{1}{e^2})$.

Summing these contributions, the initial terms $(e - \frac{1}{e})$ cancel out, leaving $(e^2 - \frac{1}{e^2}) - \frac{1}{2}(e^2 - \frac{1}{e^2}) + \frac{1}{2}(e^3 - \frac{1}{e^3})$. This further simplifies to $\frac{1}{2}(e^2 - \frac{1}{e^2}) + \frac{1}{2}(e^3 - \frac{1}{e^3})$, which combines to $\frac{1}{2} (e^{2} + e^{3} - \frac{1}{e^{2}} - \frac{1}{e^{3}})$.

The property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$ provides an immediate strategy for integrals featuring symmetric terms in the denominator. Applying this to $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1+e^{\sin x}} dx$, the integrand transforms into $32 \cos^4 x \left( \frac{1}{1+e^{\sin x}} + \frac{1}{1+e^{-\sin x}} \right)$. Since $\frac{1}{1+e^{\sin x}} + \frac{e^{\sin x}}{e^{\sin x}+1}$ equals $1$, the expression simplifies to $\int_{0}^{\frac{\pi}{4}} 32 \cos^4 x dx$.

The integral $\int_{0}^{\frac{\pi}{4}} 32 \cos^4 x dx$ simplifies by using the identity $\cos^2 x = \frac{1+\cos 2x}{2}$, resulting in $8 \int_{0}^{\frac{\pi}{4}} (1+\cos 2x)^2 dx$. Expanding this square gives $8 \int_{0}^{\frac{\pi}{4}} (1 + 2\cos 2x + \cos^2 2x) dx$, and substituting $\cos^2 2x = \frac{1+\cos 4x}{2}$ leads to $8 \int_{0}^{\frac{\pi}{4}} \left( \frac{3}{2} + 2\cos 2x + \frac{\cos 4x}{2} \right) dx$. Integration yields $8 \left[ \frac{3x}{2} + \sin 2x + \frac{\sin 4x}{8} \right]_{0}^{\frac{\pi}{4}}$, which evaluates to $8 \left( \frac{3\pi}{8} + 1 \right) = 3\pi + 8$.

The convolution integral $\int_0^x t^2 \sin(x-t) dt$ satisfies the second-order linear differential equation $I''(x) + I(x) = x^2$ obtained via Leibniz's rule, with initial conditions $I(0) = 0$ and $I'(0) = 0$. The solution to this differential equation is $I(x) = 2 \cos x + x^2 - 2$, which, when equated to the given expression $x^2$, simplifies to $2 \cos x = 2$ or $\cos x = 1$. The values of $x$ satisfying this condition in the interval $[0, 100]$ are of the form $x = 2n\pi$ for integer values of $n$. Since $0 \le 2n\pi \le 100$, the range for $n$ is $0 \le n \le \lfloor \frac{50}{\pi} \rfloor = 15$, resulting in $16$ distinct elements.

The integral identity $\cot^{-1}(z) = \tan^{-1}(\frac{1}{z})$ allows the integrand to be expressed as $\tan^{-1}\left(\frac{1}{1 + 2x(2x-1)}\right)$, which simplifies to $\tan^{-1}(2x) - \tan^{-1}(2x-1)$ using the identity $\tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1}\left(\frac{u-v}{1+uv}\right)$ with $u=2x$ and $v=2x-1$. Integrating this difference term by term, the expression becomes $4 \int_{0}^{1} \tan^{-1}(2x) dx - 4 \int_{0}^{1} \tan^{-1}(2x-1) dx$. For the first term, applying integration by parts with $U = \tan^{-1}(2x)$ and $dV = dx$ yields $[x \tan^{-1}(2x)]_0^1 - \int_0^1 \frac{2x}{1+4x^2} dx = \tan^{-1}(2) - \left[ \frac{1}{4} \ln(1+4x^2) \right]_0^1 = \tan^{-1}(2) - \frac{1}{4} \ln(5)$. The second term $\int_{0}^{1} \tan^{-1}(2x-1) dx$ evaluates to $0$ because the integrand is an odd function centered at the midpoint $x=0.5$ after the substitution $t=2x-1$. Combining these results gives $4(\tan^{-1}(2) - \frac{1}{4} \ln(5)) = 4 \tan^{-1}(2) - \ln(5)$, which implies $a=4$ and $b=1$. Therefore, $2a + b = 2(4) + 1 = 9$.

The integral is determined by piecewise splitting the interval $(-\pi/2, \pi/2)$ based on the constant values of the greatest integer functions $[\sin x]$, $[\cos x]$, and $[x]$. In the interval $[-\pi/2, 0)$, $[\sin x] = -1$ and $[\cos x] = 0$, so the denominator $3 + [\sin x] + [\cos x] = 3 - 1 + 0 = 2$. In the interval $(0, \pi/2]$, $[\sin x] = 0$ and $[\cos x] = 0$, so the denominator $3 + [\sin x] + [\cos x] = 3 + 0 + 0 = 3$.

The integral becomes $I = \int_{-\pi/2}^0 \frac{12(3+[x])}{2} dx + \int_0^{\pi/2} \frac{12(3+[x])}{3} dx = 6 \int_{-\pi/2}^0 (3+[x]) dx + 4 \int_0^{\pi/2} (3+[x]) dx$. For the first part, since $[x] = -2$ for $x \in [-\pi/2, -1)$ and $[x] = -1$ for $x \in [-1, 0)$, the value is $6 \left( \int_{-\pi/2}^{-1} (3-2) dx + \int_{-1}^0 (3-1) dx \right) = 6 \left( (\pi/2 - 1) + 2(1) \right) = 3\pi + 6$. For the second part, since $[x] = 0$ for $x \in [0, 1)$ and $[x] = 1$ for $x \in [1, \pi/2]$, the value is $4 \left( \int_0^1 (3+0) dx + \int_1^{\pi/2} (3+1) dx \right) = 4 \left( 3(1) + 4(\pi/2 - 1) \right) = 8\pi - 4$. Combining these parts yields $I = 3\pi + 6 + 8\pi - 4 = 11\pi + 2$.

Differentiating both sides of the integral equation $6\int_{1}^{x} f(t)\,dt = 3x f(x) + x^{3} - 4$ with respect to $x$ using the Fundamental Theorem of Calculus results in $6 f(x) = 3 f(x) + 3x f'(x) + 3x^{2}$. Simplifying this expression yields $3 f(x) = 3x f'(x) + 3x^{2}$, which rearranges to the form $f'(x) - \frac{f(x)}{x} = -x$.

This linear differential equation is solved using the integrating factor $e^{\int -(1/x) dx} = \frac{1}{x}$, which leads to $\frac{d}{dx} \left( \frac{f(x)}{x} \right) = -1$. Integrating both sides with respect to $x$ gives $\frac{f(x)}{x} = -x + C$, or $f(x) = -x^{2} + Cx$. Substituting $x=1$ into the original equation, $0 = 3f(1) + 1 - 4$, implies $3f(1) = 3$, so $f(1) = 1$. Using this value in $f(x) = -x^{2} + Cx$ gives $1 = -1 + C$, meaning $C = 2$.

With $f(x) = -x^{2} + 2x$, the required values are $f(2) = -(2)^{2} + 2(2) = 0$ and $f(3) = -(3)^{2} + 2(3) = -3$. Consequently, the difference $f(2) - f(3) = 0 - (-3) = 3$.

The integral $6 \int_{0}^{\pi} \left| \sin 3x + \sin 2x + \sin x \right| \, dx$ is simplified by applying the sum-to-product identity $\sin 3x + \sin x = 2 \sin 2x \cos x$, resulting in the expression $\left| \sin 2x (2 \cos x + 1) \right| = \left| 2 \sin x \cos x (2 \cos x + 1) \right|$. Given that $\sin x \ge 0$ on the interval $[0, \pi]$, the integral becomes $12 \int_{0}^{\pi} \sin x \left| \cos x (2 \cos x + 1) \right| \, dx$.

By substituting $t = \cos x$ and $dt = -\sin x \, dx$, the integral transforms into $12 \int_{-1}^{1} |t(2t+1)| \, dt$. The roots of the quadratic $2t^{2+}t$ are $t = -1/2$ and $t = 0$, dividing the interval $[-1, 1]$ into segments where the absolute value function changes sign. Evaluating the integral over these segments, we have $12 \left( \int_{-1}^{-1/2} (2t^{2+}t) \, dt - \int_{-1/2}^{0} (2t^{2+}t) \, dt + \int_{0}^{1} (2t^{2+}t) \, dt \right)$. Computing the antiderivative $\frac{2}{3}t^3 + \frac{1}{2}t^2$ at the boundaries gives $12 \left( \frac{5}{24} + \frac{1}{24} + \frac{28}{24} \right)$, which simplifies to $12 \left( \frac{34}{24} \right) = 17$.

Applying the substitution $u = \frac{tx}{36}$ with $dt = \frac{36}{x} du$ transforms the integral into $\frac{36}{x} \int_{0}^{x} f(u) du = 4\alpha f(x)$, which simplifies to $9 \int_{0}^{x} f(u) du = \alpha x f(x)$. Substituting the standard parabola $f(x) = ax^2 + bx + c$ yields $9 (\frac{ax^3}{3} + \frac{bx^2}{2} + cx) = \alpha x(ax^2 + bx + c)$, resulting in $3ax^3 + 4.5bx^2 + 9cx = \alpha ax^3 + \alpha bx^2 + \alpha cx$. Equating the coefficients of like powers of $x$ implies $\alpha = 3$, which forces $b = 0$ and $c = 0$, establishing $f(x) = ax^2$. Given $f(2) = 1$, the coefficient $a = \frac{1}{4}$, and calculating $f(-4)$ results in $\frac{1}{4}(-4)^2 = 4 = \beta$. The final value is $\beta^\alpha = 4^3 = 64$.

The greatest integer function $[y]$ satisfies $[y] = y - \{y\}$, where $0 \leq \{y\} < 1$, allowing the limit definition of $f(x)$ to be rewritten as $f(x) = \lim_{n \to \infty } \frac{1}{n^3} \sum_{k=1}^{n} \left(\frac{k^2}{3^x} - \left\{\frac{k^2}{3^x}\right\}\right)$. Since the sum $\sum_{k=1}^{n} \left\{\frac{k^2}{3^x}\right\}$ is bounded by $n$, the term $\frac{1}{n^3} \sum_{k=1}^{n} \left\{\frac{k^2}{3^x}\right\}$ approaches $0$ as $n \to \infty$, simplifying the expression to $f(x) = \frac{1}{3^x} \lim_{n \to \infty } \frac{1}{n^3} \sum_{k=1}^{n} k^2$. Using the power sum formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, the expression becomes $f(x) = \frac{1}{3^x} \lim_{n \to \infty } \frac{n(n+1)(2n+1)}{6n^3} = \frac{1}{3^x} \cdot \frac{2}{6} = \frac{1}{3^{x+1}}$.

The summation $12 \sum_{j=1}^{\infty } f(j)$ is equivalent to $12 \sum_{j=1}^{\infty } \frac{1}{3^{j+1}} = 12 \left(\frac{1}{3^2} + \frac{1}{3^3} + \dots\right)$. This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$, which sums to $\frac{a}{1-r} = \frac{1/9}{1-1/3} = \frac{1}{6}$. Multiplying by $12$ yields $12 \cdot \frac{1}{6} = 2$.

The property of an arithmetic progression dictates that the middle term is half the sum of the adjacent terms, yielding the expression $f(a) = \frac{(2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})}{2}$. Since $2^{1-a} + 2^{1+a}$ equals $2(2^{-a} + 2^a)$, the function simplifies to $f(a) = \frac{3^a + 3^{-a}}{2} + (2^a + 2^{-a})$. Given that any expression of the form $x + \frac{1}{x}$ attains a minimum value of 2 at $x=1$, we find the minimum value $\alpha$ of $f(a)$ at $a=0$, which is calculated as $\frac{2}{2} + 2 = 3$.

With $\alpha = 3$, the integral limits are defined as $\log_e(2)$ to $\log_e(3)$, leading to $I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}}$. By substituting $t = e^{2x}$, the differential term becomes $dt = 2e^{2x}dx$, or $dx = \frac{dt}{2t}$, and the limits transform to $e^{2\log_e 2} = 4$ and $e^{2\log_e 3} = 9$. The integral simplifies to $I = \int_{4}^{9} \frac{dt}{2t(t - 1/t)} = \int_{4}^{9} \frac{dt}{2(t^{2-}1)}$. Applying the standard integration formula $\int \frac{du}{u^{2-}1} = \frac{1}{2}\log_e \left|\frac{u-1}{u+1}\right|$, we obtain $\frac{1}{4} \left[ \log_e \left|\frac{t-1}{t+1}\right| \right]_{4}^{9}$, which evaluates to $\frac{1}{4} (\log_e (\frac{8}{10}) - \log_e (\frac{3}{5}))$. This results in $\frac{1}{4} \log_e (\frac{4/5}{3/5}) = \frac{1}{4} \log_e (\frac{4}{3})$.

Applying the Leibniz Integral Rule to the given integral equation $f(x) = 1 - 2x + e^x \int_{0}^{x} e^{-t} f(t) dt$ yields the derivative $f'(x) = -2 + e^x \int_{0}^{x} e^{-t} f(t) dt + e^x(e^{-x}f(x))$. Substituting the original equation's term $e^x \int_{0}^{x} e^{-t} f(t) dt = f(x) - 1 + 2x$ into this expression produces the first-order linear differential equation $f'(x) = -2 + (f(x) - 1 + 2x) + f(x)$, which simplifies to $f'(x) - 2f(x) = 2x - 3$. Multiplying by the integrating factor $e^{-2x}$ leads to $\frac{d}{dx} (f(x) e^{-2x}) = (2x - 3) e^{-2x}$. Integrating both sides gives $f(x) e^{-2x} = \int (2x - 3) e^{-2x} dx = (1-x) e^{-2x} + C$. Using the initial condition $f(0) = 1$, we find $C = 0$, confirming $f(x) = 1 - x$.

Substituting $f(x) = 1 - x$ into the expression for $g(x)$ results in $g'(x) = (f(x)+2)^{15}(x-4)^6(x+12)^{17} = (3-x)^{15}(x-4)^6(x+12)^{17}$. The critical points are $x = -12, 3, 4$. Evaluating the sign changes of $g'(x)$ shows that $g'(x)$ transitions from negative to positive at $x = -12$, identifying it as a local minimum ($p = -12$), and from positive to negative at $x = 3$, identifying it as a local maximum ($q = 3$). The point $x = 4$ is a point of inflection because the factor $(x-4)^6$ has an even exponent, causing no sign change in the derivative. Therefore, $|p + q| = |-12 + 3| = 9$.

For the function to be continuous at $x = \pi/2$, the right-hand limit of $f(x)$ as $x \to \pi/2$ must equal $1/3$. As $x$ approaches $\pi/2$, the denominator $(\pi-2x)^2$ approaches zero, which requires the numerator $b(1-\sin x)$ to vanish as well. Substituting $x = \pi/2 + h$, where $h \to 0$, transforms the expression into $\lim_{h \to 0} \frac{b(1-\cos h)}{4h^2}$. Applying the standard limit $\lim_{h \to 0} \frac{1-\cos h}{h^2} = 1/2$, the expression evaluates to $b/8$, and setting $b/8 = 1/3$ gives $b = 8/3$.

The upper bound of the integral is calculated as $3(8/3) - 6 = 2$, requiring the evaluation of $\int_{0}^{2} |x^{2+}2x-3| \, dx$. The quadratic $x^{2+}2x-3$ factors into $(x+3)(x-1)$, which remains negative for $x \in [0, 1)$ and becomes positive for $x \in (1, 2]$. By splitting the integral at $x=1$ to account for the absolute value, the calculation becomes $-\int_{0}^{1} (x^{2+}2x-3) \, dx + \int_{1}^{2} (x^{2+}2x-3) \, dx$. Evaluating the antiderivative $\frac{x^3}{3} + x^2 - 3x$ at these bounds yields $5/3$ for the first interval and $7/3$ for the second, resulting in a total value of 4.

Simplifying complex radicals often involves identifying hidden algebraic identities in the denominator, specifically recognizing that $x^{4}+x^{2}+1$ can be factored into $(x^{2}-x+1)(x^{2}+x+1)$. By applying this factorization, the expression inside the integral simplifies significantly as the term $\sqrt{x^{2}+x+1}$ cancels out from the numerator and denominator, leaving $\sqrt{\frac{x}{(x+1)(x^{2}-x+1)}}$, which further reduces to $\sqrt{\frac{x}{x^{3}+1}}$.

With the integral now expressed as $\int_{0}^{2} \frac{\sqrt{x}}{\sqrt{x^{3}+1}} dx$, a strategic substitution makes the evaluation straightforward. By setting $t=x^{3/2}$, the differential becomes $dt = \frac{3}{2}\sqrt{x}dx$, which implies that $\sqrt{x}dx = \frac{2}{3}dt$. Simultaneously, the integration limits change from $x=0$ to $t=0$, and from $x=2$ to $t=2^{3/2} = 2\sqrt{2}$.

This substitution transforms the integral into the standard form $\frac{2}{3} \int_{0}^{2\sqrt{2}} \frac{dt}{\sqrt{t^{2}+1}}$. Evaluating this integral yields $\frac{2}{3} \ln(t + \sqrt{t^{2}+1})$ evaluated from $0$ to $2\sqrt{2}$. Substituting these limits results in $\frac{2}{3} (\ln(2\sqrt{2} + \sqrt{(2\sqrt{2})^{2}+1}) - \ln(0 + \sqrt{0+1}))$, which simplifies to $\frac{2}{3} \ln(2\sqrt{2} + \sqrt{8+1})$, ultimately resulting in $\frac{2}{3} \log_{e}(3+2\sqrt{2})$.

When evaluating definite integrals over a symmetric interval $[-a, a]$, odd functions integrate to zero while even functions simplify to twice the integral over $[0, a]$. In the expression $\int_{-1}^{1} \frac{x^{3}+|x|+1}{x^{2}+2|x|+1} dx$, the denominator simplifies to $(|x| + 1)^2$ because $x^2 = |x|^2$. Dividing the numerator by this denominator splits the integral into $\int_{-1}^{1} \frac{x^3}{(|x|+1)^2} dx + \int_{-1}^{1} \frac{1}{|x|+1} dx$.

The first term is an odd function, meaning it yields zero over the symmetric range $[-1, 1]$. This leaves the second term, which is an even function, allowing it to be rewritten as $2 \int_{0}^{1} \frac{1}{x+1} dx$.

Calculating the definite integral of $\frac{1}{x+1}$ gives $\ln|x+1|$, and evaluating this from $0$ to $1$ results in $2(\ln 2 - \ln 1)$, which simplifies to $2 \log_{e} 2$.

The inverse trigonometric identity $\cot^{-1}(z) = \tan^{-1}(1/z)$ allows transforming the integrand into $\tan^{-1} \left( \frac{1}{1+x+x^{2}} \right)$, which can be expressed as $\tan^{-1} \left( \frac{1}{1+x(x+1)} \right)$. Using the property $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$ where $A=x+1$ and $B=x$, the integral simplifies to the difference $\int_{0}^{1} \tan^{-1}(x+1) dx - \int_{0}^{1} \tan^{-1}(x) dx$.

Adjusting the first integral by substituting $u=x+1$ shifts the limits to $1$ and $2$, resulting in the expression $\int_{1}^{2} \tan^{-1}(x) dx - \int_{0}^{1} \tan^{-1}(x) dx$. Applying the anti-derivative $\int \tan^{-1}(x) dx = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^{2})$ and evaluating these definite boundaries leads to the final calculation of $2 \tan^{-1} 2 - \frac{1}{2} \log_{e} \left( \frac{5}{4} \right) - \frac{\pi}{2}$.

Recognizing that the integrand originates from the product rule of differentiation significantly simplifies the problem. By differentiating the function $f(x) = \frac{\cot x}{\cos^4 x}$, we obtain $f'(x) = (-\csc^2 x)(\sec^4 x) + (\cot x)(4 \sec^3 x \cdot \sec x \tan x)$. Given that $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$, the second term in this derivative simplifies to $\frac{4}{\cos^4 x}$, resulting in $f'(x) = \frac{4 - \csc^2 x}{\cos^4 x}$. This confirms that the anti-derivative of the expression is $\frac{\cot x}{\cos^4 x}$.