Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The condition

simplifies to the differential relation $f(x)f''(x) - (f'(x))^2 = 0$. Rearranging this equation into the form $\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$ allows for integration on both sides, resulting in $\ln(f'(x)) = \ln(f(x)) + C$, which simplifies to the linear relation $f'(x) = cf(x)$. Applying the initial conditions $f(0)=1$ and $f'(0)=2$ determines the constant $c$ to be $2$, establishing $f'(x) = 2f(x)$.

Solving the separable differential equation $\frac{f'(x)}{f(x)} = 2$ involves integrating both sides to obtain $\ln(f(x)) = 2x + D$, which leads to the general solution $f(x) = ke^{2x}$. Using the condition $f(0)=1$ confirms that $k=1$, uniquely defining the function as $f(x) = e^{2x}$. Evaluating this function at $x=1$ yields $f(1) = e^2$, which is approximately $7.389$, a value that resides within the interval $(6, 9)$.

Integration by parts serves as the primary method for evaluating integrals involving products of algebraic and logarithmic functions. Applying this method to the integral $I_{n}=\int_{1}^{e} x^{19}(\ln x)^{n} dx$ by selecting $(\ln x)^{n}$ as the first function and $x^{19}$ as the second function leads to the expression $I_{n} = \left[ \frac{x^{20}}{20}(\ln x)^{n} \right]_{1}^{e} - \int_{1}^{e} \frac{x^{20}}{20} \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx$. Evaluating the boundaries at $e$ and $1$ simplifies the first term to $\frac{e^{20}}{20}$, resulting in the recurrence relation $I_{n} = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$, which can be rearranged into $20 I_{n} = e^{20} - n I_{n-1}$.

By applying this recurrence relation for $n=10$ and $n=9$, we obtain the two specific equations $20 I_{10} = e^{20} - 10 I_{9}$ and $20 I_{9} = e^{20} - 9 I_{8}$. Since both expressions equate to $e^{20}$, we set them equal to each other to get $20 I_{10} + 10 I_{9} = 20 I_{9} + 9 I_{8}$. Isolating $20 I_{10}$ on the left side yields $20 I_{10} = 10 I_{9} + 9 I_{8}$. Comparing this with the given form $20 I_{10} = \alpha I_{9} + \beta I_{8}$ identifies the values $\alpha = 10$ and $\beta = 9$, which makes $\alpha - \beta$ equal to $1$.

The symmetry property of definite integrals, which states that $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, is the key to simplifying this expression. Since the limits are symmetric about the origin, replacing $x$ with $-x$ transforms the factor $(1+e^{x\cos x})$ into $(1+e^{-x\cos x})$. By rewriting this as $(1+\frac{1}{e^{x\cos x}})$, we find that adding the integral to its transformed counterpart causes the exponential terms to cancel, resulting in $2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{\sin^4 x+\cos^4 x}$. Because the integrand is even, this simplifies to $I = \int_{0}^{\frac{\pi}{4}} \frac{dx}{\sin^4 x+\cos^4 x}$.

Dividing the numerator and denominator by $\cos^4 x$ yields $I = \int_{0}^{\frac{\pi}{4}} \frac{\sec^4 x}{\tan^4 x+1} dx$. We can rewrite the numerator as $(1+\tan^2 x)\sec^2 x$ and perform the substitution $t = \tan x - \cot x$. With this choice of $t$, the derivative becomes $dt = (\sec^2 x + \csc^2 x) dx = (1+\frac{1}{\tan^2 x})\sec^2 x dx$, which perfectly matches the numerator when the denominator is expressed as $t^{2+}2$. As $x$ ranges from $0$ to $\frac{\pi}{4}$, the variable $t$ moves from $-\infty$ to $0$. Calculating the resulting definite integral $\int_{-\infty }^{0} \frac{dt}{t^{2+}2}$ gives $\left[\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right)\right]_{-\infty }^0$, which evaluates to $\frac{1}{\sqrt{2}} \left( 0 - (-\frac{\pi}{2}) \right)$, providing a final value of $\frac{\pi}{2\sqrt{2}}$.

The Fundamental Theorem of Calculus provides the foundation for this problem by relating the derivative of the accumulation function $F(x) = \int_{0}^{x} f(t) dt$ to the integrand, establishing that $F'(x) = f(x)$. By substituting this relationship into the given integral, we can combine the terms to simplify the expression significantly.

Using the provided function $f(x) = e^{-x} \sin x$, the integral becomes:
$\int_{0}^{1} [F'(x) + f(x)] e^x dx = \int_{0}^{1} 2 f(x) e^x dx = 2 \int_{0}^{1} \sin x dx$
This calculation evaluates to $2 [-\cos x]_{0}^{1} = 2(1 - \cos 1)$. To approximate this value, we apply the Maclaurin series for $\cos x$ around zero, where $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$.

Substituting $x=1$ into the series, we find that the value of the integral is $2(1 - (1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} + \dots))$, which simplifies to $2(\frac{1}{2} - \frac{1}{24} + \frac{1}{720} - \dots)$. Taking the first two terms gives a lower bound of $1 - \frac{1}{12} = \frac{330}{360}$. Including the third term yields an upper bound of $\frac{330}{360} + \frac{1}{360} = \frac{331}{360}$. Since the series is alternating and decreasing, the actual value lies within the interval $\left[\frac{330}{360}, \frac{331}{360}\right]$.

Because $f(x) = \log_{e}\left(x+\sqrt{x^{2}+1}\right)$ is an odd function, its contribution to the integral over the symmetric interval $[-\pi/2, \pi/2]$ vanishes, simplifying the expression to $g(t) = \pi \cos\left(\frac{\pi}{4} t\right)$. Substituting $t=1$ gives $g(1) = \frac{\pi}{\sqrt{2}}$, while substituting $t=0$ gives $g(0) = \pi$. Multiplying $g(1)$ by $\sqrt{2}$ yields $\pi$, which matches $g(0)$ and confirms $\sqrt{2} g(1) = g(0)$.

Since the matrix $A$ is defined with zero entries for $i > j$, it is an upper triangular matrix, which makes its determinant equal to the product of its diagonal entries $a_{11}a_{22}a_{33}$. Each diagonal entry $a_{ii}$ involves the integral $J_{n,m} = \int_{0}^{1/2} \frac{x^n}{x^m - 1} \, dx$. For the given indices, the diagonal elements correspond to the difference $J_{i+6,3} - J_{i+3,3}$, which simplifies to $\int_{0}^{1/2} \frac{x^{i+6} - x^{i+3}}{x^3 - 1} \, dx$. Recognizing that $x^{i+6} - x^{i+3} = x^{i+3}(x^3 - 1)$, the expression reduces to the simpler integral $\int_{0}^{1/2} x^{i+3} \, dx$.

Calculating these for $i=1, 2,$ and $3$ yields $\int_{0}^{1/2} x^4 \, dx = \frac{1}{5 \cdot 2^5}$, $\int_{0}^{1/2} x^5 \, dx = \frac{1}{6 \cdot 2^6}$, and $\int_{0}^{1/2} x^6 \, dx = \frac{1}{7 \cdot 2^7}$. The determinant is therefore $|A| = \frac{1}{5 \cdot 2^5} \cdot \frac{1}{6 \cdot 2^6} \cdot \frac{1}{7 \cdot 2^7} = \frac{1}{210 \cdot 2^{18}}$. Applying the property for the determinant of an adjoint matrix, $|\operatorname{adj} M| = |M|^{n-1}$, for $M = A^{-1}$ and $n = 3$, we find that $|\operatorname{adj} A^{-1}| = |A^{-1}|^2 = \frac{1}{|A|^2}$. Substituting the value of $|A|$ results in $(210 \cdot 2^{18})^2 = (2 \cdot 105)^2 \cdot 2^{36}$, which simplifies to $105^2 \cdot 2^{38}$.

The definite integral of the polynomial $P(x) = x^2 + bx + c$ over the interval $[0, 1]$ yields the equation $\int_{0}^{1} (x^2 + bx + c) \, dx = \left[ \frac{x^3}{3} + \frac{bx^2}{2} + cx \right]_0^1 = \frac{1}{3} + \frac{b}{2} + c = 1$, which simplifies to $\frac{b}{2} + c = \frac{2}{3}$. Applying the Remainder Theorem, which states that dividing a polynomial $P(x)$ by $(x - a)$ results in a remainder equal to $P(a)$, we use the given condition $P(2) = 5$ to establish $2^2 + b(2) + c = 5$, or $2b + c = 1$.

Solving this system of linear equations involves isolating $c = 1 - 2b$ from the second equation and substituting it into the first, which results in $\frac{b}{2} + (1 - 2b) = \frac{2}{3}$. Multiplying the entire equation by 2 clears the denominators to produce $b + 2 - 4b = \frac{4}{3}$, simplifying to $-3b = -\frac{2}{3}$, and thus $b = \frac{2}{9}$. Substituting $b$ back into the expression for $c$ yields $c = 1 - 2(\frac{2}{9}) = \frac{5}{9}$. Consequently, the value of $9(b + c)$ is calculated as $9(\frac{2}{9} + \frac{5}{9}) = 9(\frac{7}{9}) = 7$.

The integral $\int_{0}^{\pi} \sin^{3} x e^{-\sin^{2} x} dx$ exploits symmetry about $x = \pi/2$, simplifying to $2 \int_{0}^{\pi/2} \sin^{3} x e^{-\sin^{2} x} dx$. Through a process of integration by parts and appropriate variable substitution, the expression is converted into the required form $\alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$. This algebraic manipulation confirms the values of the constants as $\alpha = 2$ and $\beta = 3$. Summing these values yields $\alpha + \beta = 5$.

To determine the parameter $a$, we evaluate the integral $\int_{-a}^{a} (|x| + |x-2|) \, dx = 22$ by partitioning the domain at the critical points 0 and 2. This splits the calculation into three parts: $\int_{-a}^{0} (-2x + 2) \, dx + \int_{0}^{2} 2 \, dx + \int_{2}^{a} (2x - 2) \, dx = 22$. Computing the antiderivatives and applying the limits yields $(a^2 + 2a) + 4 + (a^2 - 2a) = 22$, which simplifies to $2a^2 = 18$, confirming $a = 3$. With $a$ identified, the required integral is $\int_{3}^{-3} (x + [x]) \, dx$. Given that $x$ is an odd function, its integral over the symmetric interval $[-3, 3]$ is $0$. The integral of the greatest integer function $[x]$ over the same range is computed by summing the constant value of $[x]$ on each sub-interval, which equals $-3$. Since the integral is defined from 3 to -3, we take the negative of this sum, resulting in 3.

The functional equation $f(x) + f(x+1) = 2$ implies that the function averages to $1$ over any interval of length $1$. By integrating this relation over an arbitrary interval, one can deduce that the integral of $f(x)$ over any interval of length $2$ is constantly $2$. This property provides the foundation for simplifying the given integrals $I_1 = \int_{0}^{8} f(x) dx$ and $I_2 = \int_{-1}^{3} f(x) dx$.

The expression $I_1 + 2I_2$ simplifies by evaluating each component separately. Since the integral over any interval of length $2$ is $2$, we can partition $I_1 = \int_{0}^{8} f(x) dx$ into four intervals of length $2$, resulting in $I_1 = 4 \times 2 = 4 \int_{0}^{2} f(x) dx$. Similarly, we can transform $I_2 = \int_{-1}^{3} f(x) dx$ by substituting $x$ with $u-1$ or shifting the bounds, which yields $\int_{0}^{4} f(x+1) dx$. Using the original identity $f(x+1) = 2 - f(x)$, this integral becomes $\int_{0}^{4} (2 - f(x)) dx$, which evaluates to $8 - 2 \int_{0}^{2} f(x) dx$. Multiplying this entire expression by $2$ gives $2I_2 = 16 - 4 \int_{0}^{2} f(x) dx$. Summing these two components, $I_1 + 2I_2$ results in $4 \int_{0}^{2} f(x) dx + 16 - 4 \int_{0}^{2} f(x) dx$, which equals $16$.

Isolating the real part of a complex reciprocal involves multiplying the numerator and denominator by the complex conjugate of the denominator, which is $(1 - \cos \theta) - 2i \sin \theta$. The denominator becomes $(1 - \cos \theta)^2 + (2 \sin \theta)^2$, and by applying the half-angle identities $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$, this expression simplifies to $4 \sin^2 \frac{\theta}{2} (\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2})$. Since the numerator is $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$, the real part is $\frac{1}{2 (\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2})}$.

Setting this equal to $\frac{1}{5}$ implies that $2 (\sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2}) = 5$. By replacing $\sin^2 \frac{\theta}{2}$ with $1 - \cos^2 \frac{\theta}{2}$, the equation becomes $2 (1 - \cos^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2}) = 5$, or $2 (1 + 3 \cos^2 \frac{\theta}{2}) = 5$. This simplifies to $3 \cos^2 \frac{\theta}{2} = \frac{3}{2}$, leading to $\cos^2 \frac{\theta}{2} = \frac{1}{2}$. Given that $\theta \in (0, \pi)$, we find $\frac{\theta}{2} = \frac{\pi}{4}$, which yields $\theta = \frac{\pi}{2}$.

Evaluating the definite integral $\int_{0}^{\pi/2} \sin x dx$ results in $[-\cos x]_{0}^{\pi/2}$. Substituting the limits gives $-\cos(\frac{\pi}{2}) - (-\cos 0)$, which evaluates to $0 - (-1)$, resulting in a value of $1$.

The absolute value function $|\sin 2x|$ exhibits a periodic nature where the curve repeats every $\pi/2$ interval, which implies that the total integral over the range $[0, \pi]$ is simply twice the area found within the first interval $[0, \pi/2]$. Because $\sin 2x$ remains positive throughout this specific range, the absolute value signs can be removed, simplifying the problem to the calculation of $2 \int_{0}^{\pi/2} \sin 2x dx$.

Evaluating the antiderivative of $\sin 2x$ yields $-\frac{\cos 2x}{2}$, which must be calculated across the specified boundaries. Substituting $\pi/2$ and $0$ into this expression leads to $-\frac{1}{2}(\cos \pi - \cos 0)$. Given that $\cos \pi$ is $-1$ and $\cos 0$ is $1$, the bracketed expression simplifies to $-1 - 1$, which is $-2$. Multiplying $-\frac{1}{2}$ by $-2$ results in $1$, and doubling this value to account for the entire range of the original integral provides a final answer of $2$.

The Fundamental Theorem of Calculus allows us to determine the derivative of the curve by identifying it as the integrand of the function $y(x)=\int_{0}^{x}(2 t^{2}-15 t+10) d t$, which gives $y'(x)=2 x^{2}-15 x+10$. Since the normal to the curve at $(a, b)$ is parallel to the line $x+3y=-5$, the slope of the normal must match the slope of this line, which is $-\frac{1}{3}$. Because the slope of a normal is the negative reciprocal of the tangent slope, we set $-\frac{1}{y'(a)} = -\frac{1}{3}$, establishing that the tangent slope $y'(a)$ must be equal to $3$.

Equating the derivative expression to this value results in the quadratic equation:
$2 a^{2}-15 a+10=3$
This simplifies to $2 a^{2}-15 a+7=0$, which factors as $(2 a-1)(a-7)=0$, yielding potential solutions $a=\frac{1}{2}$ and $a=7$. Given the condition $a>1$, we must discard the first solution and proceed with $a=7$. We then find the ordinate $b$ by evaluating the definite integral at $x=7$:
$b = \int_{0}^{7}(2 t^{2}-15 t+10) d t = \left[ \frac{2}{3}t^{3} - \frac{15}{2}t^{2} + 10t \right]_0^7 = -\frac{413}{6}$
Substituting $a=7$ and $b=-\frac{413}{6}$ into the expression $|a+6b|$ yields $|7+6(-\frac{413}{6})|$, which simplifies to $|7-413|$, resulting in a final value of 406.

This problem is a classic application of the Riemann sum definition of a definite integral, where the limit of a summation transforms into an integration problem as the number of terms approaches infinity. To identify the underlying pattern, first rewrite the leading term $\frac{1}{n}$ as $\frac{n}{n^2}$, which allows the entire expression to be written as the summation $\sum_{r=0}^{n-1} \frac{n}{(n+r)^2}$. Dividing both the numerator and the denominator inside the sum by $n^2$ reveals the form $\sum_{r=0}^{n-1} \frac{1}{n} \cdot \frac{1}{(1 + \frac{r}{n})^2}$.

By definition, as $n$ tends to infinity, the ratio $\frac{r}{n}$ behaves like the variable $x$, and the fraction $\frac{1}{n}$ acts as the differential $dx$. Since the index $r$ ranges from $0$ to $n-1$, the bounds for the definite integral are $0$ and $1$. This effectively converts the limit into the integral $\int_{0}^{1} \frac{1}{(1+x)^2} \, dx$. Computing this integral using the standard antiderivative formula results \in the calculation $\left[ -\frac{1}{1+x} \right]_{0}^{1} = -\left( \frac{1}{2} - 1 \right)$, which simplifies to $\frac{1}{2}$.

The greatest integer function $[x]$ acts as a constant over every interval of length one, which simplifies the integration process significantly. We partition the range from $0$ to $10$ into segments $[n, n+1]$ where the expression $[x] e^{[x]}$ simplifies to $n e^n$, allowing us to compute the integral as a summation of these individual segments. This transforms the problem into $\sum_{n=0}^{9} \int_{n}^{n+1} \frac{n e^n}{e^{x-1}} dx$.

Moving the constant factors $n e^n$ and the $e$ from the denominator outside the integral sign, each segment becomes $n e^{n+1} \int_{n}^{n+1} e^{-x} dx$. Applying the fundamental theorem of calculus, the integral of $e^{-x}$ evaluates to $-(e^{-(n+1)} - e^{-n})$, or $e^{-n} - e^{-(n+1)}$. When we multiply this result by $n e^{n+1}$, the terms simplify elegantly: $n e^{n+1} (e^{-n} - e^{-(n+1)}) = n(e^1 - e^0) = n(e-1)$.

Since the term for $n=0$ is zero, we sum $n(e-1)$ for integers $n$ from $1$ to $9$. Factoring out the $(e-1)$, we are left with the arithmetic series $(e-1) \sum_{n=1}^{9} n$. Applying the formula for the sum of the first nine positive integers, $\frac{9 \times 10}{2}$, we arrive at $45(e-1)$.

The greatest integer function $[f(x)]$ maintains a constant value between intervals where $f(x)$ is an integer. To evaluate \int_{1}^{3} \[x^{2-}2x-2]$ , dx$, start by completing the square to rewrite the integrand as$[(x-1)^{2-}3]$. Since$3$is an integer, it can be removed from the floor function to become$[(x-1)^2]-3$, allowing the integral to be split into$\int_{1}^{3} $[(x-1)^2]$ , dx - \int_{1}^{3} 3 , dx$. Using the substitution$t = x-1$, the limits shift to$0$and$2$, transforming the expression into$\int_{0}^{2} $[t^2]$ , dt - 6$.

With the integral simplified to \int_{0}^{2} \[t^2]$ , dt$, we identify the points where$[t^2]$changes its value, which occur when$t^2$is an integer, specifically at$t=1$,$t=\sqrt{2}$, and$t=\sqrt{3}$. This divides the range$[0, 2]$into four intervals:$[0, 1]$,$[1, \sqrt{2}]$,$[\sqrt{2}, \sqrt{3}]$, and$[\sqrt{3}, 2]$. On these segments,$[t^2]$takes the constant values$0, 1, 2,$and$3$respectively. Calculating the area for each segment gives$0(1-0) + 1(\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3})$, which simplifies to$\sqrt{2}-1 + 2\sqrt{3}-2\sqrt{2} + 6-3\sqrt{3}$, or$5-\sqrt{2}-\sqrt{3}$. Subtracting the original$6$from this sum yields$-\sqrt{2}-\sqrt{3}-1$.

The sum $f(e) + f\left(\frac{1}{e}\right)$ relies on the definition $f(x) = \int_{1}^{x} \frac{\log_e t}{1+t} \, dt$. Expressing the sum as two separate integrals gives $f(e) = \int_{1}^{e} \frac{\log_e t}{1+t} \, dt$ and $f\left(\frac{1}{e}\right) = \int_{1}^{1/e} \frac{\log_e t}{1+t} \, dt$. To evaluate $f\left(\frac{1}{e}\right)$, apply the substitution $t = \frac{1}{u}$, which yields $dt = -\frac{1}{u^2} \, du$. As $t$ ranges from $1$ to $\frac{1}{e}$, $u$ changes from $1$ to $e$. Substituting these into the integral results in $-\int_{1}^{e} \frac{\log_e(1/u)}{1+1/u} \cdot \frac{1}{u^2} \, du$. Simplifying the integrand gives $-\int_{1}^{e} \frac{-\log_e u}{\frac{u+1}{u}} \cdot \frac{1}{u^2} \, du$, which resolves to $\int_{1}^{e} \frac{\log_e u}{u(1+u)} \, du$.

Combining the two expressions, $f(e) + f\left(\frac{1}{e}\right) = \int_{1}^{e} \frac{\log_e t}{1+t} \, dt + \int_{1}^{e} \frac{\log_e t}{t(1+t)} \, dt$. Factoring out $\log_e t$ allows the integrand to be written as $\log_e t \left( \frac{1}{1+t} + \frac{1}{t(1+t)} \right)$. Simplifying the terms inside the parentheses results in $\frac{t+1}{t(1+t)}$, which reduces to $\frac{1}{t}$. The sum simplifies to the integral $\int_{1}^{e} \frac{\log_e t}{t} \, dt$. By using the substitution $v = \log_e t$, where $dv = \frac{1}{t} \, dt$, the integral transforms into $\int_{0}^{1} v \, dv$. Evaluating this yields $\left[ \frac{v^2}{2} \right]_{0}^{1} = \frac{1}{2}$.