Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Continuity and differentiability at a specific point require analyzing the function's limits and derivatives from both directions. For $x > 2$, the integral expression $\int_{0}^{x}(5+|1-t|)\,dt$ can be split at $t=1$ as $\int_{0}^{1}(5+(1-t))\,dt + \int_{1}^{x}(5+(t-1))\,dt$, which evaluates to the polynomial $f(x) = \frac{x^2}{2} + 4x + 1$. Substituting $x=2$ into this expression yields $2+8+1=11$, matching the value $5(2)+1=11$ obtained from the piece defined for $x \le 2$, which confirms the function is continuous. Checking differentiability, the derivative of the lower piece is $5$, while the derivative of the upper piece is $\frac{d}{dx}(\frac{x^2}{2} + 4x + 1) = x+4$; evaluated at $x=2$, these yield $5$ and $6$, respectively. Because these one-sided derivatives do not match, the function is not differentiable at $x=2$, whereas the integral of a continuous function ensures differentiability at $x=1$.

Differentiating the integral equation $f(x)=\int_{0}^{x} e^{t} f(t) d t+e^{x}$ using the Leibniz rule yields the first-order differential equation $f'(x)=e^{x} f(x)+e^{x}$ Factoring the right side gives $f'(x)=e^x(f(x)+1)$, which rearranges to the separable form $\frac{f'(x)}{f(x)+1}=e^x$

Integrating both sides results in $\ln(f(x)+1)=e^x+C$. Setting $x=0$ in the original equation gives $f(0)=1$, so $\ln(1+1)=e^{0+}C$, which determines the constant $C=\ln(2)-1$. Substituting this constant back into the logarithmic expression and exponentiating both sides leads to $f(x)+1=e^{e^x+\ln(2)-1}$, which simplifies to the final form $f(x)=2e^{e^x-1}-1$

The greatest integer function $\lfloor x^{3} \rfloor$ remains constant over specific intervals, necessitating the division of the integral at points where $x^3$ crosses an integer. Throughout the interval $(-1, 0)$, $x^3$ takes values between $-1$ and $0$, which means $\lfloor x^{3} \rfloor = -1$. In the interval $(0, 1)$, $x^3$ falls between $0$ and $1$, leading to $\lfloor x^{3} \rfloor = 0$.

Splitting the integral $\int_{-1}^{1} x^{2} e^{\lfloor x^{3} \rfloor} \, dx$ into these two regions results in the sum $\int_{-1}^{0} x^{2} e^{-1} \, dx + \int_{0}^{1} x^{2} e^{0} \, dx$. This expression simplifies to $\frac{1}{e} \int_{-1}^{0} x^{2} \, dx + \int_{0}^{1} x^{2} \, dx$. Applying the power rule to integrate $x^2$ gives $\frac{1}{e} \left[ \frac{x^{3}}{3} \right]_{-1}^{0} + \left[ \frac{x^{3}}{3} \right]_{0}^{1}$. Substituting the boundaries yields $\frac{1}{e} \left( 0 - \left( -\frac{1}{3} \right) \right) + \left( \frac{1}{3} - 0 \right)$, which resolves to $\frac{1}{3e} + \frac{1}{3}$, ultimately combining to $\frac{e+1}{3e}$.

Evaluating the integral $I_n = \int_{\pi/4}^{\pi/2} \cot^n x \, dx$ begins by splitting the integrand into $\cot^{n-2} x \cot^2 x$. Using the trigonometric identity $\cot^2 x = \csc^2 x - 1$, the expression becomes $I_n = \int_{\pi/4}^{\pi/2} \cot^{n-2} x (\csc^2 x - 1) \, dx$. This algebraic manipulation allows us to rearrange the terms and isolate the sum $I_n + I_{n-2} = \int_{\pi/4}^{\pi/2} \cot^{n-2} x \csc^2 x \, dx$.

Applying the substitution $u = \cot x$ implies $du = -\csc^2 x \, dx$, which transforms the integral boundaries from $[\pi/4, \pi/2]$ to $[1, 0]$. Computing the resulting integral $\int_1^0 -u^{n-2} \, du$ is equivalent to $\int_0^1 u^{n-2} \, du$, yielding the result $I_n + I_{n-2} = \frac{1}{n-1}$. Testing this formula with $n=4, 5, 6$ produces $I_4 + I_2 = \frac{1}{3}$, $I_5 + I_3 = \frac{1}{4}$, and $I_6 + I_4 = \frac{1}{5}$. Because the reciprocals $3, 4, 5$ form a sequence with a constant common difference of 1, the terms $\frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}$ form an arithmetic progression.

This problem relies on the principle that the limit of a Riemann sum as $n$ approaches infinity can be expressed as a definite integral. By dividing both the numerator and the denominator of the fraction within the summation by $n^2$, the general term transforms into $\frac{1}{1+4(\frac{r}{n})^2}$. This allows the entire limit to be rewritten in the standard integral form $\int_{0}^{2} \frac{1}{1+4x^2} dx$, where the variable $x$ represents the ratio $\frac{r}{n}$ and the limits span from 0 to 2. Applying the integral rule for $\frac{1}{1+(ax)^2}$, which is $\frac{1}{a} \tan^{-1}(ax)$, leads to the evaluation of $\frac{1}{2} \tan^{-1}(2x)$ at the boundaries 0 and 2. Computing the difference between these boundaries results in the final value $\frac{1}{2} \tan^{-1}(4)$.

King's property of definite integrals states that for any function $f(x)$ on the interval $[a, b]$, the integral $\int_{a}^{b} f(x) dx$ is equal to $\int_{a}^{b} f(a+b-x) dx$. Letting $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\lfloor x \rfloor + \lfloor -\sin x \rfloor) dx$, we apply this property with $a+b = 0$ to express the integral as $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\lfloor -x \rfloor + \lfloor \sin x \rfloor) dx$.

Combining these two representations allows us to write $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} ((\lfloor x \rfloor + \lfloor -x \rfloor) + (\lfloor \sin x \rfloor + \lfloor -\sin x \rfloor)) dx$. Using the identity that for any non-integer value $t$, the sum $\lfloor t \rfloor + \lfloor -t \rfloor$ is always $-1$, and noting that the isolated points where $x$ or $\sin x$ are integers have no effect on the value of the definite integral, the integrand simplifies to $-1 + (-1) = -2$. The integral expression becomes $2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (-2) dx$, which evaluates to $-2 \times (\frac{\pi}{2} - (-\frac{\pi}{2})) = -2\pi$. Dividing both sides by two, we find $I = -\pi$.

The evaluation relies on the definition of the Beta function $I_{m,n} = \int_0^1 x^{m-1}(1-x)^{n-1} dx$ By applying the substitution $x = \frac{1}{t+1}$, which results \in $dx = -\frac{1}{(t+1)^2} dt$, the limits shift to $(0, ∞)$ and the integral transforms into $I_{m,n} = \int_0^\infty \frac{t^{n-1}}{(t+1)^{m+n}} dt$ Since $I_{m,n}$ is symmetric with respect to $m$ and $n$, it can also be written as $I_{m,n} = \int_0^∞ \frac{t^{m-1}}{(t+1)^{m+n}} dt$, and adding these two equivalent forms gives $2 I_{m,n} = \int_0^\infty \frac{t^{n-1} + t^{m-1}}{(t+1)^{m+n}} dt$

Splitting this integral at the point $t=1$ yields the sum of two integrals, one from $0$ to $1$ and the other from $1$ to $\infty$. Substituting $t = \frac{1}{z}$ into the second integral, where $dt = -\frac{1}{z^2} dz$, transforms it into the same expression as the first integral over the range $(0, 1)$. This indicates that the two parts are identical, leading to $2 I_{m,n} = 2 \int_0^1 \frac{t^{m-1} + t^{n-1}}{(t+1)^{m+n}} dt$ Dividing both sides by $2$ demonstrates that the target integral equals $I_{m,n}$, confirming that $\alpha = 1$.

Calculating the value of this definite integral requires partitioning the interval $[-1/2, 1]$ into three distinct regions based on the piecewise nature of $\lfloor 2x \rfloor$ and $|x|$. For the interval $-1/2 \le x < 0$, the floor function $\lfloor 2x \rfloor$ evaluates to $-1$ and the absolute value function $|x|$ simplifies to $-x$, yielding an integrand of $-1 - x$. In the range $0 \le x < 1/2$, the floor function equals $0$ while $|x|$ simplifies to $x$, resulting in an integrand of $x$. For the final portion $1/2 \le x \le 1$, the floor function is $1$ and $|x|$ is $x$, making the integrand $1 + x$.

Integrating these parts individually, the first section $\int_{-1/2}^{0} (-1 - x) \, dx$ leads to $\left[ -x - \frac{x^2}{2} \right]_{-1/2}^{0}$, which equals $-3/8$. The second section $\int_{0}^{1/2} x \, dx$ results in $\left[ \frac{x^2}{2} \right]_{0}^{1/2}$, or $1/8$. The third section $\int_{1/2}^{1} (1 + x) \, dx$ becomes $\left[ x + \frac{x^2}{2} \right]_{1/2}^{1}$, which evaluates to $7/8$. Adding these values, $-3/8 + 1/8 + 7/8$, gives a total area of $5/8$. Multiplying this result by $8$ provides the final value of $5$.

The integral of a function over a fixed interval remains unchanged if the variable $x$ is replaced by the sum of the limits minus $x$, expressed as $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. For the given expression, observe that the term $x^{2-}44x+484$ is equivalent to $(22-x)^2$. Applying logarithm rules to handle the exponents, the integrand simplifies to $\frac{2 \log_e x}{2 \log_e x + 2 \log_e(22-x)}$, which further reduces to $\frac{\log_e x}{\log_e x + \log_e(22-x)}$.

Given the lower limit $6$ and upper limit $16$, their sum is $22$, so replacing $x$ with $22-x$ transforms the integral into $I = \int_{6}^{16} \frac{\log_e(22-x)}{\log_e(22-x) + \log_e x} \, dx$. Adding the original integral to this transformed version yields $2I = \int_{6}^{16} \frac{\log_e x + \log_e(22-x)}{\log_e x + \log_e(22-x)} \, dx$, which equates to $\int_{6}^{16} 1 \, dx$. Calculating the difference between the upper and lower limits results in $16-6=10$, meaning $2I=10$, which leaves $I=5$.

The function $f(x)=\int_{0}^{x}[y]dy$ represents the accumulated area under a step function, which naturally creates a continuous piecewise linear graph where each segment joins perfectly at the integer bounds. By evaluating the integral as $f(x) = \sum_{k=0}^{n-1} k + \int_{n}^{x} n dy = \frac{n(n-1)}{2} + n(x-n)$ for $x \in [n, n+1)$, we see that the value at any integer $n$ is defined by $\frac{n(n-1)}{2}$, confirming that the left and right limits are identical at every integer point. However, because the integrand $[y]$ experiences a jump discontinuity of size 1 at every integer, the derivative of $f(x)$, which is $[x]$, jumps from $n-1$ to $n$ at these locations. This sudden change in the slope of the function signifies that $f$ is differentiable everywhere except at the integer points.

The greatest integer function $[x]$ behaves as a step function, which requires splitting the integral into separate intervals where the function remains constant. For the range $0 \le x < 1$, the value of $[x]$ is $0$, which simplifies the expression $(x-[x])^{[x]}$ to $x^0$, or $1$. This reduces the first portion of the integral to $\pi^{2} \int_{0}^{1} \sin\left(\frac{\pi x}{2}\right) dx$. Integrating $\sin\left(\frac{\pi x}{2}\right)$ results in $-\frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right)$, and evaluating this from $0$ to $1$ yields $\frac{2}{\pi}$. Multiplying this result by $\pi^{2}$ gives $2\pi$.

For the range $1 \le x < 2$, the value of $[x]$ is $1$, which makes the expression $(x-[x])^{[x]}$ equal to $(x-1)^1$. The remaining part of the integral becomes $\pi^{2} \int_{1}^{2} (x-1) \sin\left(\frac{\pi x}{2}\right) dx$. Using integration by parts with $u = x-1$ and $dv = \sin\left(\frac{\pi x}{2}\right) dx$, the integral is calculated as $\pi^{2} \left[ (x-1) \left(-\frac{2}{\pi} \cos \frac{\pi x}{2}\right) \right]_{1}^{2} + \pi^{2} \int_{1}^{2} \frac{2}{\pi} \cos\left(\frac{\pi x}{2}\right) dx$. Evaluating the boundary terms at $1$ and $2$ yields $2\pi$, while integrating the remaining trigonometric component results in $-4$. Combining the two parts of the entire expression gives $2\pi + 2\pi - 4$, which simplifies to $4(\pi - 1)$.

Multiplying the equation by $e^x$ yields $e^x F(x) = \int_{3}^{x} (3t^2 + 2t + 4F'(t)) dt$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus introduces the derivative term, giving $e^x F(x) + e^x F'(x) = 3x^2 + 2x + 4F'(x)$. Regrouping this equation results in $(e^x - 4)F'(x) + e^x F(x) = 3x^2 + 2x$, which is the derivative of the product $F(x)(e^x - 4)$.

Integrating both sides with respect to $x$ provides $F(x)(e^x - 4) = x^3 + x^2 + C$. Setting $x=3$ and knowing that $F(3)=0$ determines the constant $C = -36$, which leads to the explicit form $F(x) = \frac{x^3 + x^2 - 36}{e^x - 4}$. To find the derivative $F'(x)$, apply the quotient rule to obtain $\frac{(3x^2 + 2x)(e^x - 4) - (x^3 + x^2 - 36)e^x}{(e^x - 4)^2}$.

Substituting $x=4$ into this expression yields $\frac{(3(16) + 2(4))(e^4 - 4) - (64 + 16 - 36)e^4}{(e^4 - 4)^2}$. This simplifies to $\frac{56(e^4 - 4) - 44e^4}{(e^4 - 4)^2}$, which further reduces to $\frac{12e^4 - 224}{(e^4 - 4)^2}$. Comparing this expression with $\frac{\alpha e^{\beta} - 224}{(e^{\beta} - 4)^2}$ identifies $\alpha = 12$ and $\beta = 4$, so the value of $\alpha + \beta$ is 16.

The Fundamental Theorem of Calculus allows us to relate integrals to their underlying functions by differentiating both sides of the given equation with respect to $x$. When we differentiate the integral $\int_{0}^{x} \sqrt{1-(f'(t))^{2}} dt = \int_{0}^{x} f(t) dt$, we obtain $\sqrt{1-(f'(x))^{2}} = f(x)$, which simplifies upon squaring to $1-(f'(x))^{2} = (f(x))^{2}$. Rearranging this expression yields $(f'(x))^{2} = 1-(f(x))^{2}$, leading to the separable differential equation $f'(x) = \sqrt{1-(f(x))^{2}}$ since the function is non-negative.

Integrating both sides gives $\int \frac{df}{\sqrt{1-f^{2}}} = \int dx$, resulting in $\arcsin(f(x)) = x + C$. Given the initial condition $f(0)=0$, we find that $\arcsin(0) = 0 + C$, which confirms $C=0$ and thus $f(x) = \sin(x)$. Evaluating the final limit $\lim_{x\rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) dt$ requires substituting $f(t) = \sin(t)$ and applying L'Hôpital's Rule to the indeterminate form $\frac{\int_{0}^{x} \sin(t) dt}{x^{2}}$, which yields $\lim_{x\rightarrow 0} \frac{\sin(x)}{2x}$, resulting in a final value of $1/2$.

This problem relies on a fundamental property of definite integrals, which states that replacing $x$ with the sum of the limits minus $x$ leaves the value of the integral unchanged. Specifically, for an integral over the interval $[a, b]$, the function behaves symmetrically such that $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. In this case, the sum of the limits $\frac{\pi}{24} + \frac{5\pi}{24}$ equals $\frac{\pi}{4}$, so the transformation involves substituting $x$ with $\frac{\pi}{4} - x$.

Rewriting the integrand $\frac{1}{1+(\tan 2x)^{1/3}}$ as $\frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}$ makes this substitution straightforward. When $x$ is replaced by $\frac{\pi}{4} - x$, the term $2x$ becomes $\frac{\pi}{2} - 2x$. Using trigonometric identities, the cosine and sine functions swap values, turning the integrand into $\frac{(\sin 2x)^{1/3}}{(\sin 2x)^{1/3} + (\cos 2x)^{1/3}}$. Since adding the original integral to this transformed version results in the integral of $1$ over the same interval, the value of the original integral is simply half the width of the interval. The difference between the limits is $\frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24}$, or $\frac{\pi}{6}$, which means the value of the integral is half of this, resulting in $\frac{\pi}{12}$.

The periodicity of the sine-squared function allows the integral over $[0, 100\pi]$ to be decomposed into 100 identical intervals of $[0, \pi]$, effectively transforming the problem into $100 \int_{0}^{\pi} e^{-x/\pi} \sin^2 x \, dx$. Utilizing the trigonometric identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ allows us to simplify the expression to $50 \int_{0}^{\pi} e^{-x/\pi} (1 - \cos 2x) \, dx$. This enables us to split the problem into two parts, $I_{1} = \int_{0}^{\pi} e^{-x/\pi} \, dx$ and $I_{2} = \int_{0}^{\pi} e^{-x/\pi} \cos 2x \, dx$.

Evaluating $I_{1}$ yields $\pi (1 - e^{-1})$, while $I_{2}$ requires integration by parts, which reveals that $I_{2} = \frac{\pi (1 - e^{-1})}{1 + 4\pi^{2}}$. Combining these parts, the total value becomes $50 \left( \pi (1 - e^{-1}) - \frac{\pi (1 - e^{-1})}{1 + 4\pi^{2}} \right)$. Simplifying this expression leads to $\frac{200 \pi^3 (1 - e^{-1})}{1 + 4\pi^{2}}$. By matching this result to the form $\frac{\alpha \pi^3}{1 + 4\pi^{2}}$, it is evident that $\alpha$ equals $200 (1 - e^{-1})$.

The integral of an absolute value function necessitates splitting the domain at the roots where the inner expression changes sign, as the function behavior dictates the sign of the integrand. Factoring the quadratic expression $3x^2 - 3x - 6$ yields $3(x-2)(x+1)$, which reveals that the function is positive on the interval $[-2, -1]$ and negative on the interval $[-1, 2]$, allowing the absolute value to be removed by splitting the integration path into these two specific segments. Calculating the sum $3\int_{-2}^{-1} (x^2 - x - 2) dx + 3\int_{-1}^{2} (-x^2 + x + 2) dx$ using the antiderivative $\frac{x^3}{3} - \frac{x^2}{2} - 2x$ leads to the final value of 19.

Recognizing the expression inside the square root as a perfect square is the key to solving this integral, as it follows the form $(a-b)^2 = a^2 + b^2 - 2ab$ where $a = \frac{x+1}{x-1}$, $b = \frac{x-1}{x+1}$, and the product $ab=1$. The integrand simplifies to $\left| \frac{x+1}{x-1} - \frac{x-1}{x+1} \right|$, which algebraically reduces to $\left| \frac{4x}{x^{2-}1} \right|$. Leveraging the symmetry of the interval $[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$ and the even nature of the function, we can compute $2 \int_0^{\frac{1}{\sqrt{2}}} \left| \frac{4x}{x^{2-}1} \right| \, dx$.

Within the positive interval $[0, \frac{1}{\sqrt{2}}]$, the term $x^{2-}1$ is negative, so the absolute value requires a sign flip, resulting in $-8 \int_0^{\frac{1}{\sqrt{2}}} \frac{x}{x^{2-}1} \, dx$, or equivalently $-4 \int_0^{\frac{1}{\sqrt{2}}} \frac{2x}{x^{2-}1} \, dx$. Integrating this leads to -4 \[ \ln |x^{2-}1| ]_0^{\frac{1}{\sqrt{2}}}$, which evaluates to$-4 ( \ln | \frac{1}{2} - 1 | - \ln | -1 | )$. Since$\ln(1/2) = -\ln 2$, the final value is$4 \ln 2$, which is identical to$\log_e 16$.

Determining the interval $(a, b)$ where the function exists requires ensuring all nested logarithmic arguments remain positive. This mandates the condition $\log_5(\log_3(18x-x^{2-}77)) > 0$, implying that $\log_3(18x-x^{2-}77) > 1$, which further simplifies to $18x-x^{2-}77 > 3$. Rearranging this inequality into $x^{2-}18x+80 < 0$ and factoring the quadratic results in $(x-8)(x-10) < 0$, which defines the domain as $x \in (8, 10)$. This establishes the integral limits $a=8$ and $b=10$.

The integral $I = \int_{8}^{10} \frac{\sin^3 x}{\sin^3 x + \sin^3(18-x)} \, dx$ utilizes the definite integral property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. Applying this transformation to the integrand yields $I = \int_{8}^{10} \frac{\sin^3(18-x)}{\sin^3(18-x) + \sin^3 x} \, dx$. Summing these two expressions for $I$ results in:
$2I = \int_{8}^{10} \frac{\sin^3 x + \sin^3(18-x)}{\sin^3 x + \sin^3(18-x)} \, dx = \int_{8}^{10} 1 \, dx$
Evaluating this integral yields $2I = 10 - 8 = 2$, which simplifies to $I = 1$.

Evaluating $\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x} \, dx$ is straightforward using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$, which simplifies the integrand when the limits of integration sum to zero. Replacing $x$ with $-x$ transforms the integrand into $\frac{\cos^2(-x)}{1+3^{-x}}$, which simplifies to $\frac{3^x \cos^2 x}{3^x+1}$ because $\cos(-x) = \cos x$. If we define the original integral as $I$ and add it to this transformed version, the terms combine to give $2I = \int_{-\pi/2}^{\pi/2} \frac{(1+3^x) \cos^2 x}{1+3^x} \, dx$, which simplifies to $\int_{-\pi/2}^{\pi/2} \cos^2 x \, dx$. Substituting $\cos^2 x$ with $\frac{1+\cos 2x}{2}$ and integrating leads to a result of $\frac{\pi}{2}$, meaning $I = \frac{\pi}{4}$.