Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The rationalization of the integrand provides the most direct path to solving the integral. By multiplying the numerator and denominator by the conjugate $\sqrt{3+x^2} - \sqrt{1+x^2}$, the integrand simplifies from $\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}$ to $\frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2}$. Incorporating the constant factor of 4, the expression transforms into $2 \int_{0}^{1} \sqrt{3+x^2} \, dx - 2 \int_{0}^{1} \sqrt{1+x^2} \, dx$.

To evaluate these integrals, utilize the standard form $\int \sqrt{a^{2+}x^2} \, dx = \frac{x}{2}\sqrt{a^{2+}x^2} + \frac{a^2}{2} \ln(x+\sqrt{a^{2+}x^2})$. For the first term, applying the limits from 0 to 1 yields $\left[ \frac{x}{2}\sqrt{3+x^2} + \frac{3}{2}\ln(x+\sqrt{3+x^2}) \right]_{0}^{1}$, which evaluates to $1 + \frac{3}{2}\ln(3) - \frac{3}{2}\ln(\sqrt{3})$, simplified to $1 + \frac{3}{2}\ln(\sqrt{3})$. Similarly, the second term $\int_{0}^{1} \sqrt{1+x^2} \, dx$ evaluates to $\left[ \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\ln(x+\sqrt{1+x^2}) \right]_{0}^{1}$, resulting in $\frac{\sqrt{2}}{2} + \frac{1}{2}\ln(1+\sqrt{2})$.

Combining these results, the integral becomes $2 \left( 1 + \frac{3}{2}\ln(\sqrt{3}) - \frac{\sqrt{2}}{2} - \frac{1}{2}\ln(1+\sqrt{2}) \right)$, which simplifies to $2 + 3\ln(\sqrt{3}) - \sqrt{2} - \ln(1+\sqrt{2})$. Subtracting the final term $3 \ln(\sqrt{3})$ from this total isolates the required value as $2 - \sqrt{2} - \ln(1+\sqrt{2})$.

The domain of a composite logarithmic function is determined by requiring each nested argument to be strictly positive, proceeding from the outermost to the innermost logarithm. For $f(x) = \log_2(\log_4(\log_6(3+4x-x^2)))$, the function is defined when $\log_4(\log_6(3+4x-x^2)) > 0$, which requires $\log_6(3+4x-x^2) > 1$, and subsequently $3+4x-x^2 > 6$. This simplifies to the inequality $x^{2-}4x+3 < 0$, which factors as $(x-1)(x-3) < 0$, identifying the domain interval $(a, b)$ as $(1, 3)$ where $a=1$ and $b=3$.

Evaluating the integral \int_0^2 \[x^2]$ dx$requires observing that the greatest integer function$[x^2]$equals zero for$x \in [0, 1)$, allowing us to focus on the interval$[1, 2]$. We partition the integral at the points where$x^2$is an integer:$\int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^2 3 dx$. Computing these parts gives$(\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3})$, which simplifies to$5 - \sqrt{2} - \sqrt{3}$. Comparing this result to$p-\sqrt{q}-\sqrt{r}$yields$p=5$,$q=2$, and$r=3$, leading to the final sum$p+q+r = 10$.

The intersection points of the curve and the line define the integral bounds, while the symmetric property of the definite integral allows for the simplification of the integrand to solve for the final value. Solving the system $x^2 = 2y$ and $y - 2x - 6 = 0$ by substituting $y = \frac{x^2}{2}$ leads to $x^2 - 4x - 12 = 0$, which factors as $(x-6)(x+2) = 0$, identifying the second quadrant intersection point at $(-2, 2)$ where the limits are $a = -2$ and $b = 2$. Evaluating the integral $I = \int_{-2}^{2} \frac{9x^2}{1+5^x} dx$ is efficient using the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$, as summing the integrand with its counterpart replacing $x$ with $-x$ simplifies the term to $9x^2$. Thus, $2I = \int_{-2}^{2} 9x^2 dx$ reduces to $I = \int_{0}^{2} 9x^2 dx$, and calculating $9 \left[ \frac{x^3}{3} \right]_0^2$ yields 24.

To compute the integral of an absolute value, the strategy requires splitting the definite integral at the points where the function inside the modulus changes sign. Within the interval $[-1, \frac{3}{2}]$, the integrand $\pi^2 x \sin(\pi x)$ changes sign at $x=0$ and $x=1$. Because the function $f(x) = x \sin(\pi x)$ is even, the integral from $-1$ to $0$ is identical to the integral from $0$ to $1$, allowing the entire expression to be simplified as $2 \pi^2 \int_{0}^{1} x \sin(\pi x) \, dx - \pi^2 \int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx$, where the negative sign on the second term accounts for the function being negative on that interval.

Evaluating the indefinite integral $\int x \sin(\pi x) \, dx$ via integration by parts, with $u=x$ and $dv=\sin(\pi x) \, dx$, yields the expression $-\frac{x \cos(\pi x)}{\pi} + \frac{\sin(\pi x)}{\pi^2}$. Applying the limits of integration, the definite integral $\int_{0}^{1} x \sin(\pi x) \, dx$ evaluates to $1/\pi$, while $\int_{1}^{\frac{3}{2}} x \sin(\pi x) \, dx$ results in $-\frac{1}{\pi^2} - \frac{1}{\pi}$. Multiplying these values by their respective coefficients produces $2 \pi^2 (1/\pi) - \pi^2 (-\frac{1}{\pi^2} - \frac{1}{\pi})$, which simplifies to $2\pi + 1 + \pi$, yielding a final sum of $3\pi + 1$.

The substitution $u = \sin \theta - \cos \theta$ serves as an efficient method because its derivative, $(\cos \theta + \sin \theta) d\theta$, appears directly in the numerator, while the trigonometric term $\sin 2\theta$ can be expressed as $1 - u^2$ since $u^2 = 1 - \sin 2\theta$. Substituting these into the integral, the denominator becomes $9 + 16(1 - u^2)$, which simplifies to $25 - 16u^2$. Adjusting the limits of integration, $\theta = 0$ corresponds to $u = -1$, and $\theta = \pi/4$ corresponds to $u = 0$, resulting in the integral $\int_{-1}^{0} \frac{du}{25 - 16u^2}$.

Factoring out $16$ from the denominator yields the form $\frac{1}{16} \int_{-1}^{0} \frac{du}{(5/4)^2 - u^2}$, and applying the standard integral formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right|$ with $a = 5/4$ results in $\frac{1}{16} \cdot \frac{1}{2(5/4)} \ln \left| \frac{5/4 + u}{5/4 - u} \right|_{-1}^{0}$, which simplifies to $\frac{1}{40} \ln \left| \frac{5+4u}{5-4u} \right|_{-1}^{0}$. Evaluating this expression at the limits $u=0$ and $u=-1$ yields $\frac{1}{40} [ \ln(1) - \ln(1/9) ] = \frac{\ln 9}{40}$. Multiplying the result by $80$ produces $80 \cdot \frac{\ln 9}{40} = 2 \ln 9$, which simplifies to $4 \ln 3$ because $9 = 3^2$.

Differentiating the given integral equation with respect to $x$ transforms it into a first-order linear differential equation. Applying the Fundamental Theorem of Calculus to the integral and the product rule to the term involving $f(x)$ yields $10 f(x) = 5 f(x) + 5x f'(x) - 5x^4$. Simplifying this relationship results in $f'(x) - \frac{1}{x} f(x) = x^3$, which is solved using the integrating factor $e^{\int -\frac{1}{x} dx} = \frac{1}{x}$. Multiplying the differential equation by this factor allows us to write it as $\frac{d}{dx} \left( \frac{f(x)}{x} \right) = x^2$. Integrating both sides with respect to $x$ produces $\frac{f(x)}{x} = \frac{x^3}{3} + C$, leading to the general form $f(x) = \frac{x^4}{3} + Cx$.

To determine the constant $C$, we evaluate the original integral equation at $x=1$, which gives $10(0) = 5(1)f(1) - 1 - 9$, resulting in $f(1) = 2$. Substituting $x=1$ into our expression for $f(x)$ yields $2 = \frac{1}{3} + C$, so $C = \frac{5}{3}$. With the function fully defined as $f(x) = \frac{x^4}{3} + \frac{5}{3}x$, we calculate $f(3) = \frac{81}{3} + \frac{5}{3}(3) = 27 + 5 = 32$.

The relationship between the integral function $g(x) = \int_0^x t f(t) dt$ and its integrand $f(t)$ is defined by the derivative $g'(x) = x f(x)$, meaning $f(x) = \frac{g'(x)}{x}$. With the given $g(x^3) = x^6 + x^7$, the substitution $u = x^3$ transforms the equation into $g(u) = u^2 + u^{7/3}$. Differentiating this with respect to $u$ gives $g'(u) = 2u + \frac{7}{3}u^{4/3}$, and dividing this expression by $u$ yields $f(u) = 2 + \frac{7}{3}u^{1/3}$, which simplifies to $f(r^3) = 2 + \frac{7}{3}r$ when $u = r^3$. Evaluating the summation $\sum_{r=1}^{15} (2 + \frac{7}{3}r)$ requires calculating $2(15) + \frac{7}{3} \left( \frac{15 \times 16}{2} \right)$, which simplifies to the equation $30 + 280 = 310$.

Applying integration by parts is the most effective way to untangle these definite integrals and isolate the required values. For the integral involving the second derivative, $\int_{0}^{2} x^2 F''(x) dx = 40$, we set $u = x^2$ and $dv = F''(x) dx$, which gives $du = 2x dx$ and $v = F'(x)$. This transforms the expression into $[x^2 F'(x)]_{0}^{2} - \int_{0}^{2} 2x F'(x) dx = 40$. Since we are given that $\int_{0}^{2} x F'(x) dx = 6$, the second part of the equation becomes $2(6) = 12$. Substituting this in, we get $4 F'(2) - 12 = 40$, which simplifies to $4 F'(2) = 52$, confirming that $F'(2) = 13$.

To find the value of the integral $\int_{0}^{2} F(x) dx$, we apply integration by parts to the first given equation, $\int_{0}^{2} x F'(x) dx = 6$, using $u = x$ and $dv = F'(x) dx$. This results in $[x F(x)]_{0}^{2} - \int_{0}^{2} F(x) dx = 6$. Given that $F(x) = x f(x)$ and $f(2) = 1$, we calculate the boundary term at the upper limit as $2 F(2) = 2(2 \cdot 1) = 4$. Substituting this value yields $4 - \int_{0}^{2} F(x) dx = 6$, implying $\int_{0}^{2} F(x) dx = -2$. Combining these results, $F'(2) + \int_{0}^{2} F(x) dx = 13 - 2 = 11$.

The primary tool for evaluating these integrals is the King's property, which states that for any integrable function, the integral $\int_a^b f(x) \, dx$ remains unchanged if we replace $x$ with $a+b-x$. For the integral $I_1$ with limits from $-\frac{1}{2}$ to $1$, the sum of the limits is $\frac{1}{2}$. Applying this property, we replace $x$ with $\frac{1}{2}-x$, transforming the integral into $I_1 = \int_{-\frac{1}{2}}^{1} 2\left(\frac{1}{2}-x\right) f\left(2\left(\frac{1}{2}-x\right)\left(1-2\left(\frac{1}{2}-x\right)\right)\right) \, dx$. This expression simplifies to $I_1 = \int_{-\frac{1}{2}}^{1} (1-2x) f((1-2x)(2x)) \, dx$.

By adding this transformed version of $I_1$ to the original expression for $I_1$, we get $2I_1 = \int_{-\frac{1}{2}}^{1} (2x + 1 - 2x) f(2x(1-2x)) \, dx$, which simplifies to $2I_1 = \int_{-\frac{1}{2}}^{1} f(2x(1-2x)) \, dx$. To express this in terms of $I_2$, we use the substitution $u = 2x$, which gives $du = 2 \, dx$ or $dx = \frac{du}{2}$. Adjusting the integration limits, when $x = -\frac{1}{2}$, $u = -1$, and when $x = 1$, $u = 2$. Substituting these into our equation yields $2I_1 = \int_{-1}^{2} f(u(1-u)) \frac{du}{2}$, which further simplifies to $4I_1 = \int_{-1}^{2} f(u(1-u)) \, du$. Since the right-hand side is equivalent to the definition of $I_2$, it follows that $4I_1 = I_2$, meaning the ratio $\frac{I_2}{I_1}$ is $4$.

Factoring the function $f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x$ into $(7\tan^6 x - 3\tan^2 x)(1 + \tan^2 x)$ simplifies it to $(7\tan^6 x - 3\tan^2 x) \sec^2 x$. This is a powerful step because $\sec^2 x$ is the derivative of $\tan x$, which makes the subsequent integration straightforward. For $I_1 = \int_0^{\pi/4} f(x) dx$, we use the substitution $u = \tan x$, which transforms the integral into $\int_0^1 (7u^6 - 3u^2) du$. Calculating the antiderivative $u^7 - u^3$ from $0$ to $1$ results in $1 - 1 = 0$.

For $I_2 = \int_0^{\pi/4} x f(x) dx$, we utilize integration by parts with $u = x$ and $dv = f(x) dx$, where $v = \tan^7 x - \tan^3 x$. Applying the formula $\int u dv = uv - \int v du$ yields $[x(\tan^7 x - \tan^3 x)]_0^{\pi/4} - \int_0^{\pi/4} (\tan^7 x - \tan^3 x) dx$. The boundary term evaluates to zero since $\tan(\pi/4) = 1$ and $\tan(0) = 0$. The remaining integral term is computed by recognizing the equivalent form $\int_0^{\pi/4} (\tan^3 x - \tan^5 x) \sec^2 x dx$, which results in evaluating $\frac{\tan^4 x}{4} - \frac{\tan^6 x}{6}$ from $0$ to $\pi/4$. Substituting the limits leads to $\frac{1}{4} - \frac{1}{6} = \frac{1}{12}$. Consequently, the total value $7 I_1 + 12 I_2$ evaluates to $7(0) + 12(\frac{1}{12})$, which equals $1$.

To determine the value of $f(x)$, we start with the given functional equation $f(x) + 2f(\frac{1}{x}) = x^2 + 5$. Replacing $x$ with $\frac{1}{x}$ yields $f(\frac{1}{x}) + 2f(x) = \frac{1}{x^2} + 5$. Multiplying the second equation by $2$ gives $4f(x) + 2f(\frac{1}{x}) = \frac{2}{x^2} + 10$. Subtracting the original equation from this result leads to $3f(x) = \frac{2}{x^2} - x^2 + 5$, or $f(x) = \frac{1}{3}\left(\frac{2}{x^2} - x^2 + 5\right)$. Evaluating the integral $\alpha = \int_{1}^{2} f(x) \, dx$ gives $\frac{1}{3} \int_{1}^{2} (\frac{2}{x^2} - x^2 + 5) \, dx = \frac{1}{3} \left[ -\frac{2}{x} - \frac{x^3}{3} + 5x \right]_{1}^{2} = \frac{1}{3} \left( (-1 - \frac{8}{3} + 10) - (-2 - \frac{1}{3} + 5) \right) = \frac{1}{3} (\frac{19}{3} - \frac{8}{3}) = \frac{11}{9}$, so $9\alpha = 11$.

For the function $g(x)$, the equation $2g(x) - 3g(\frac{1}{2}) = x$ implies $g(x) = \frac{x + 3g(0.5)}{2}$, which takes the linear form $g(x) = ax + b$. Substituting this into the equation $2(ax + b) - 3(0.5a + b) = x$ allows us to compare coefficients: $2a = 1$ leads to $a = 0.5$, and $2b - 1.5a - 3b = 0$ leads to $-b = 1.5(0.5) = 0.75$, so $b = -0.75$. The function is $g(x) = 0.5x - 0.75$. Calculating the integral $\beta = \int_{1}^{2} (0.5x - 0.75) \, dx$ yields $[0.25x^2 - 0.75x]_{1}^{2} = (1 - 1.5) - (0.25 - 0.75) = -0.5 - (-0.5) = 0$. Finally, the value $9\alpha + \beta = 11 + 0 = 11$.

The Beta function identity $I(m,n) = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx = \frac{(m-1)!(n-1)!}{(m+n-1)!}$ transforms the given integral expression into a factorial calculation where $I(9,14)$ becomes $\frac{8!13!}{22!}$ and $I(10,13)$ becomes $\frac{9!12!}{22!}$. Combining these terms over the common denominator $22!$ reveals a shared factor of $8!12!$ in the numerator, which simplifies the expression to $\frac{8!12!(13+9)}{22!}$. Since $13+9$ equals $22$, the expression reduces to $\frac{8!12! \cdot 22}{22!}$, which simplifies to $\frac{8!12!}{21!}$ because $22! = 22 \cdot 21!$. This final value is equivalent to $I(9,13)$, which is derived by substituting $m=9$ and $n=13$ into the original Beta function formula.

Simplifying the integrand through trigonometric division allows for an immediate substitution that reduces the problem to a basic power rule integration. Factoring $\cos^3 x$ out of the squared term in the denominator yields $\cos^6 x (1 + \tan^3 x)^2$, which transforms the original integral into $\int_{0}^{\pi/4} \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} \, dx$ because the numerator simplifies to $\frac{\sin^2 x}{\cos^4 x}$, which is equivalent to $\tan^2 x \sec^2 x$.

By defining $t = 1 + \tan^3 x$, the differential becomes $dt = 3 \tan^2 x \sec^2 x \, dx$, which simplifies the integral's numerator to $\frac{1}{3} dt$. Adjusting the limits of integration from $x \in [0, \pi/4]$ to $t \in [1, 2]$ transforms the expression into $\frac{1}{3} \int_1^2 t^{-2} \, dt$. Evaluating this definite integral results in $\frac{1}{3} [-\frac{1}{t}]_1^2$, which equals $\frac{1}{3} (1 - \frac{1}{2})$, giving a final value of $\frac{1}{6}$.

Expressing the numerator as a linear combination of the denominator and its derivative is an efficient strategy for solving integrals involving a ratio of trigonometric functions. By setting $136 \sin x = A(3 \sin x + 5 \cos x) + B(3 \cos x - 5 \sin x)$, we match the coefficients for $\sin x$ and $\cos x$ to obtain the system $3A - 5B = 136$ and $5A + 3B = 0$. Solving for $B$ in terms of $A$ gives $B = -\frac{5}{3}A$, which, when substituted into the first equation, leads to $3A + \frac{25}{3}A = 136$, resulting in $A = 12$ and $B = -20$.

This substitution transforms the integral into two parts: $\int_{0}^{\pi/4} 12 \, dx + \int_{0}^{\pi/4} \frac{-20(3 \cos x - 5 \sin x)}{3 \sin x + 5 \cos x} \, dx$. The first part results in $12 \times \frac{\pi}{4}$, which simplifies to $3\pi$. The second part evaluates to $-20 \left[ \ln(3 \sin x + 5 \cos x) \right]_0^{\pi/4}$. Applying the limits gives $-20 \left( \ln(\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{2}}) - \ln(5) \right)$, which simplifies to $-20 (\ln(4\sqrt{2}) - \ln(5))$. Because $4\sqrt{2}$ equals $2^{5/2}$, the expression becomes $-20 (\frac{5}{2} \ln 2 - \ln 5)$, resulting in $-50 \ln 2 + 20 \ln 5$. Combining both terms, the total value is $3\pi - 50 \ln 2 + 20 \ln 5$.

Since the limit evaluates to an indeterminate $\frac{0}{0}$ form as $x$ approaches $\frac{\pi}{2}$, we apply L'Hôpital's rule and the Leibniz integral rule to differentiate the numerator. Differentiating the integral $\int_{x^3}^{(\pi/2)^3} (\sin(2t^{1/3}) + \cos(t^{1/3})) dt$ with respect to $x$ yields $-( \sin(2x) + \cos(x) ) \cdot 3x^2$, while the derivative of the denominator $(x - \frac{\pi}{2})^2$ is $2(x - \frac{\pi}{2})$. This results in the expression $\lim_{x \to \frac{\pi}{2}} \frac{-( \sin(2x) + \cos(x) ) \cdot 3x^2}{2(x - \frac{\pi}{2})}$.

By utilizing the trigonometric identities $\cos(x) = \sin(\frac{\pi}{2} - x)$ and $\sin(2x) = 2\sin(x)\cos(x)$, the numerator term $-\sin(2x) - \cos(x)$ becomes $-\sin(\frac{\pi}{2} - x)(2\sin(x) + 1)$. This allows us to isolate the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, where $\theta = \frac{\pi}{2} - x$. Specifically, noting that $\frac{\sin(\frac{\pi}{2} - x)}{x - \frac{\pi}{2}} = -1$, the expression simplifies to $\frac{2\sin(\frac{\pi}{2}) + 1}{2} \cdot 3(\frac{\pi}{2})^2$. Substituting $\sin(\frac{\pi}{2}) = 1$ leads to $\frac{3}{2} \cdot 3 \cdot \frac{\pi^2}{4}$, which gives $\frac{9\pi^2}{8}$.

Recognizing that $1-\sin 2x$ can be expressed as $(\sin x - \cos x)^2$ simplifies the integrand to $\sqrt{(\sin x - \cos x)^2}$, which is equivalent to $|\sin x - \cos x|$. Since the sign of the expression inside the absolute value shifts at $\pi/4$, the integral must be evaluated in two segments: from $\pi/6$ to $\pi/4$, where $\cos x > \sin x$, and from $\pi/4$ to $\pi/3$, where $\sin x > \cos x$.

The integral becomes $\int_{\pi/6}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/3} (\sin x - \cos x) \, dx$. Evaluating these parts yields $(\sin x + \cos x)_{\pi/6}^{\pi/4} + (-\cos x - \sin x)_{\pi/4}^{\pi/3}$, which simplifies to $(\sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}) + (\sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2})$. This sum results in $2\sqrt{2} - 1 - \sqrt{3}$, identifying the constants as $\alpha = -1$, $\beta = 2$, and $\gamma = -1$. Calculating $3\alpha + 4\beta - \gamma$ with these values gives $3(-1) + 4(2) - (-1)$, which equals $6$.