Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The Fundamental Theorem of Calculus allows us to evaluate the integral by identifying the antiderivative of $3t^2 - 2 \phi'(t)$, which is $t^3 - 2 \phi(t)$. Applying the limits 5 and $x$ to this result leads to the equation $x \phi(x) = x^3 - 125 - 2 \phi(x) + 2 \phi(5)$.

The constant $\phi(5)$ is determined by substituting $x = 0$ and the given condition $\phi(0) = 4$ into the derived equation, which yields $0 = -125 - 2(4) + 2 \phi(5)$, or $2 \phi(5) = 133$. Substituting $x = 2$ into the main equation gives $2 \phi(2) = 2^3 - 125 - 2 \phi(2) + 133$. Simplifying this expression results in $4 \phi(2) = 16$, confirming that $\phi(2) = 4$.

The derivative of a cubic polynomial $P(x)$ is a quadratic function that vanishes at the local extremum points $x=1$ and $x=-1$, implying $P'(x) = a(x-1)(x+1) = a(x^2 - 1)$ for some constant $a$. Integrating this expression with respect to $x$ gives $P(x) = a(\frac{x^3}{3} - x) + C$, where $C$ is the constant of integration. We apply the integral condition $\int_{-1}^{1} P(x) \, dx = 18$ to isolate $C$. Because the integral of the odd function component $a(\frac{x^3}{3} - x)$ over the symmetric interval $[-1, 1]$ is zero, the condition reduces to $\int_{-1}^{1} C \, dx = 18$, which simplifies to $2C = 18$, resulting in $C = 9$.

Next, we use the root condition $P(-3) = 0$ to solve for $a$. Substituting $x = -3$ into our expression for $P(x)$ yields $a(\frac{(-3)^3}{3} - (-3)) + 9 = 0$, which simplifies to $a(-9 + 3) + 9 = 0$, or $-6a = -9$, giving $a = \frac{3}{2}$. Because the sum of all coefficients of any polynomial $P(x)$ is equivalent to evaluating the function at $x = 1$, we calculate $P(1) = \frac{3}{2}(\frac{1^3}{3} - 1) + 9$. Evaluating this expression gives $\frac{3}{2}(-\frac{2}{3}) + 9 = -1 + 9$, which results in a sum of 8.

Partitioning the range of integration is essential for integrals involving the greatest integer function, as the expression $[x^2]$ must remain constant within each interval. Given the integration limits from 0 to $\sqrt{\frac{\pi}{2}}$, the value of the greatest integer function changes when $x^2$ reaches the integer 1, which occurs at $x=1$.

The integral is split into two distinct parts based on this change. In the first interval from 0 to 1, $x^2$ stays between 0 and 1, making $[x^2] = 0$, so this part simplifies to $\int_{0}^{1} -\cos x \, dx$. For the second interval from 1 to $\sqrt{\frac{\pi}{2}}$, $x^2$ lies between 1 and $\frac{\pi}{2}$, resulting in $[x^2] = 1$, which transforms the integral into $\int_{1}^{\sqrt{\frac{\pi}{2}}} (1 - \cos x) \, dx$. Combining these evaluations results in a net value of $-1$, and applying the absolute value gives a final result of 1.

The greatest integer function requires partitioning the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} dx$ into five distinct unit intervals, specifically $[k, k+1]$ for $k = 0, 1, 2, 3, 4$, because the value of $[x]$ remains constant at $k$ across each interval. Within any such interval, the integrand simplifies to $\frac{x+k}{e^{x-k}}$, which can be written as $(x+k)e^{-(x-k)}$.

Substituting $t = x - k$ transforms each component integral into $\int_{0}^{1} (t+2k)e^{-t} dt$. This further decomposes into $\int_{0}^{1} t e^{-t} dt + 2k \int_{0}^{1} e^{-t} dt$. Solving these standard integrals, $\int_{0}^{1} t e^{-t} dt$ evaluates to $1 - 2e^{-1}$ using integration by parts, and $\int_{0}^{1} e^{-t} dt$ evaluates to $1 - e^{-1}$.

Summing these contributions across all five intervals results in $5(1 - 2e^{-1}) + 2(0+1+2+3+4)(1 - e^{-1})$. The sum of the integers $0$ through $4$ is $10$, so the second term becomes $20(1 - e^{-1})$. Expanding this yields $5 - 10e^{-1} + 20 - 20e^{-1}$, which simplifies to $25 - 30e^{-1}$. Comparing this to $\alpha e^{-1} + \beta$ identifies $\alpha = -30$ and $\beta = 25$, which satisfies the given condition $5\alpha + 6\beta = 0$. Calculating the final value $(\alpha + \beta)^2$ gives $(-30+25)^2$, which equals 25.

The relationship between the derivative of $f(x)$ and $f(2-x)$ allows us to establish a fixed sum for the function. By integrating both sides of $f'(x) = f'(2-x)$, we obtain $f(x) = -f(2-x) + C$. We determine the constant $C$ by substituting $x=0$, which gives $f(0) = -f(2) + C$, leading to $1 = -e^2 + C$ and therefore $C = 1 + e^2$. This implies that the sum $f(x) + f(2-x)$ is constant and equal to $1 + e^2$ throughout the interval $[0, 2]$.

Utilizing the property of definite integrals where $\int_0^2 f(x) \, dx = \int_0^2 f(2-x) \, dx$, we can denote the target integral as $I$ and represent it in two equivalent ways. Adding these two representations of the integral results in $2I = \int_0^2 [f(x) + f(2-x)] \, dx$, which simplifies to $2I = \int_0^2 (1 + e^2) \, dx$. Evaluating this integral over the interval from $0$ to $2$ gives $2I = 2(1 + e^2)$, confirming that $I = 1 + e^2$.

The evaluation of this definite integral relies on the properties of symmetric functions and the integral identity $\int_{a}^{b} h(x) \, dx = \int_{a}^{b} h(a+b-x) \, dx$. Since $f(x^2)$ is an even function, the integral from $-4$ to $4$ simplifies to $2 \int_{0}^{4} f(x^2) \, dx$, and substituting $f(x^2) = 4x^3 - g(4-x)$ results in $8 \int_{0}^{4} x^3 \, dx - 2 \int_{0}^{4} g(4-x) \, dx$. Because the integral identity implies $\int_{0}^{4} g(4-x) \, dx = \int_{0}^{4} g(x) \, dx$, combining these two terms leads to $\int_{0}^{4} (g(4-x) + g(x)) \, dx$, which is zero according to the given condition $g(4-x) + g(x) = 0$. This eliminates the function $g$ entirely, leaving the final evaluation:
$8 \int_{0}^{4} x^3 \, dx = 8 \left[ \frac{x^4}{4} \right]_{0}^{4} = 2(256) = 512$

The monotonicity property of definite integrals allows us to determine the total range of $g(3)$ by separately integrating the lower and upper bounds of $f(t)$ over its specified intervals. Since the expression $g(3) = \int_{0}^{3} f(t) dt$ is additive, we evaluate the integral by splitting it into two parts covering $[0, 1]$ and $[1, 3]$ to match the piecewise constraints given for $f(t)$.

On the first interval $[0, 1]$, the function satisfies $\frac{1}{3} \le f(t) \le 1$, which implies that $\int_{0}^{1} f(t) dt$ must fall between $\int_{0}^{1} \frac{1}{3} dt = \frac{1}{3}$ and $\int_{0}^{1} 1 dt = 1$. Similarly, for the interval $[1, 3]$, the function is bounded by $0 \le f(t) \le \frac{1}{2}$, so the integral $\int_{1}^{3} f(t) dt$ ranges from $\int_{1}^{3} 0 dt = 0$ to $\int_{1}^{3} \frac{1}{2} dt = 1$. Adding the lower bounds and upper bounds from both intervals yields the minimum and maximum possible values for the total integral, resulting in the range $\left[\frac{1}{3}, 2\right]$.

The integrand $e^{x - \lfloor x \rfloor}$ is a periodic function with a period of 1, where the integral over any interval of length 1 simplifies to $\int_{0}^{1} e^x \, dx = e - 1$. Integrating from 0 to $a$ involves summing this result over the $n = \lfloor a \rfloor$ complete intervals and adding the integral over the remaining partial interval from $n$ to $a$. This approach yields $n(e - 1) + \int_{n}^{a} e^{x - n} \, dx = 10e - 9$, which expands to $n(e - 1) + (e^{a - n} - 1) = 10e - 9$. Distributing the terms results in $ne - n + e^{a - n} - 1 = 10e - 9$, or equivalently, $ne + e^{a - n} = 10e + n - 8$. By comparing the integer and exponential components, we identify $n = 10$ and $e^{a - 10} = 2$, which confirms that $a = 10 + \log_e 2$.

The expression $e^{x-[x]}$ simplifies to $e^{\{x\}}$, where $\{x\}$ is the fractional part of $x$. Because the function $[\sin 2\pi x]$ repeats every integer interval and the fractional part function shares this periodicity, the integrand itself is periodic with period 1. The integral from 0 to 10 is therefore equivalent to 10 times the integral over the single interval $[0, 1]$, which can be written as $10 \int_0^1 \frac{[\sin 2\pi x]}{e^x} \, dx$ because $\{x\} = x$ within this range.

The floor function $[\sin 2\pi x]$ behaves predictably within $[0, 1]$. For $0 \le x < \frac{1}{2}$, the argument $2\pi x$ ranges from $0$ to $\pi$, which means $\sin 2\pi x$ resides in $[0, 1)$ and forces $[\sin 2\pi x]$ to be 0. For $\frac{1}{2} \le x < 1$, the argument $2\pi x$ ranges from $\pi$ to $2\pi$, so $\sin 2\pi x$ resides in $[-1, 0)$, resulting in $[\sin 2\pi x] = -1$. This simplifies the entire expression to $-10 \int_{\frac{1}{2}}^1 e^{-x} \, dx$.

Computing the antiderivative of $e^{-x}$ results in $-e^{-x}$. Applying the limits from $\frac{1}{2}$ to 1 yields $-10(-e^{-1} - (-e^{-\frac{1}{2}}))$, which simplifies to $10e^{-1} - 10e^{-\frac{1}{2}} + 0$. Comparing this result to the form $\alpha e^{-1} + \beta e^{-\frac{1}{2}} + \gamma$, the coefficients are $\alpha = 10$, $\beta = -10$, and $\gamma = 0$. Summing these values gives $\alpha + \beta + \gamma = 0$.

Definite integrals over symmetric intervals simplify significantly when the integrand exhibits symmetry. If a function satisfies $f(-x) = -f(x)$, it is classified as an odd function, and the integral of any such function over a symmetric range $[-a, a]$ is always zero because the area on the negative side of the y-axis perfectly cancels the area on the positive side.

For the expression
$\int_{-1}^{1} \log(x+\sqrt{x^{2}+1}) \, dx$
we examine $f(x) = \log(x+\sqrt{x^{2}+1})$ by substituting $-x$. This yields $f(-x) = \log(-x+\sqrt{(-x)^{2+}1})$, which simplifies to $\log(\sqrt{x^{2+}1}-x)$. Since $\sqrt{x^{2+}1}-x$ is the reciprocal of $x+\sqrt{x^{2+}1}$, we can rewrite the expression as $\log(1 / (x+\sqrt{x^{2+}1}))$, resulting in $-\log(x+\sqrt{x^{2+}1})$. Because $f(-x) = -f(x)$, the function is confirmed to be odd, and the resulting value of the integral is 0.

The expression $x-[x]$ represents the fractional part of $x$, a function that repeats every interval of length 1, making the integral of $e^{x-[x]}$ constant over any unit interval. By performing a substitution such as $u = x - (n-1)$, the definite integral over each interval $[n-1, n]$ reduces to $\int_{0}^{1} e^u du$, which evaluates to $e^1 - e^0$, or $e-1$. Summing this result across the 100 specified unit intervals requires multiplying the single-interval value by 100, leading to $100(e-1)$.

Recognizing the limit of a sum as a definite integral relies on transforming the expression into terms involving $j/n$ and $1/n$. By dividing both the numerator and the denominator inside the summation by $n$, the term $\frac{(2j-1)+8n}{(2j-1)+4n}$ becomes $\frac{2(j/n) - (1/n) + 8}{2(j/n) - (1/n) + 4}$. As $n$ grows arbitrarily large, the fraction $j/n$ behaves as the continuous variable $x$ and the term $1/n$ approaches zero, effectively mapping the summation to a definite integral over the interval $[0, 1]$.

This process transforms the limit into the integral:
$\int_{0}^{1} \frac{2x+8}{2x+4} \, dx$
Rewriting the integrand as $1 + \frac{4}{2x+4}$ allows for straightforward integration across the unit interval. Computing the antiderivative of $1$ results in $1$, and integrating $\frac{4}{2x+4}$ yields $2 \log_e|2x+4|$. Evaluating this expression at the boundaries $0$ and $1$ gives $1 + 2(\log_e(6) - \log_e(4))$, which simplifies to $1 + 2 \log_e\left(\frac{3}{2}\right)$.

Rewriting the integrand by expressing $x^2$ as $(x^2 - 1) + 1$ allows us to split the integral into $-\int_{0}^{1/2} \frac{1}{\sqrt{1-x^2}} dx + \int_{0}^{1/2} \frac{1}{(1-x^2)^{3/2}} dx$. The first part simplifies to -\[\sin^{-1}(x)]{0}^{1/2}$, which evaluates to$-\frac{\pi}{6}$. For the second part, applying the substitution$x = \sin\theta$with$dx = \cos\theta d\theta$changes the limits of integration to$0$and$\pi/6$, yielding$\int{0}^{\pi/6} \frac{\cos\theta}{\cos^3\theta} d\theta = \int_{0}^{\pi/6} \sec^2\theta d\theta$. Computing this integral gives$\tan(\pi/6) = \frac{1}{\sqrt{3}}$. Combining these results gives the total value as$\frac{1}{\sqrt{3}} - \frac{\pi}{6}$. Since this total equals$\frac{k}{6}$, we solve for$k$by multiplying the entire expression by$6$, leading to$k = \frac{6}{\sqrt{3}} - \pi$, which simplifies to$2\sqrt{3} - \pi$.

To solve for $\theta$, first express the equation $2\cot^2\theta - \frac{5}{\sin\theta} + 4 = 0$ entirely in terms of $\sin\theta$. Substituting the identity $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta} = \frac{1-\sin^2\theta}{\sin^2\theta}$ transforms the expression into $2\left(\frac{1-\sin^2\theta}{\sin^2\theta}\right) - \frac{5}{\sin\theta} + 4 = 0$. Multiplying the entire equation by $\sin^2\theta$ yields $2(1-\sin^2\theta) - 5\sin\theta + 4\sin^2\theta = 0$, which simplifies to the quadratic form $2\sin^2\theta - 5\sin\theta + 2 = 0$. Factoring this quadratic results in $(2\sin\theta - 1)(\sin\theta - 2) = 0$, providing solutions where $\sin\theta = \frac{1}{2}$ or $\sin\theta = 2$. Since $\sin\theta$ cannot exceed 1, the only valid solutions correspond to $\sin\theta = \frac{1}{2}$.

In the specified interval $(0, 2\pi)$, $\sin\theta = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$. Thus, the smallest value is $\theta_1 = \frac{\pi}{6}$ and the largest value is $\theta_2 = \frac{5\pi}{6}$.

With these bounds, the definite integral $\int_{\pi/6}^{5\pi/6} \cos^2 3\theta \, d\theta$ can be evaluated by applying the double-angle identity $\cos^2 x = \frac{1+\cos 2x}{2}$. Replacing the integrand with $\frac{1+\cos 6\theta}{2}$, the integral becomes:
$\int_{\pi/6}^{5\pi/6} \frac{1+\cos 6\theta}{2} \, d\theta = \frac{1}{2} \left[ \theta + \frac{\sin 6\theta}{6} \right]_{\pi/6}^{5\pi/6}$
Substituting the limits, the term $\frac{\sin 6\theta}{6}$ becomes $\frac{\sin 5\pi}{6} = 0$ and $\frac{\sin \pi}{6} = 0$, leaving only the difference in the $\theta$ terms:
$\frac{1}{2} \left( \frac{5\pi}{6} - \frac{\pi}{6} \right) = \frac{1}{2} \left( \frac{4\pi}{6} \right) = \frac{\pi}{3}$

The function $|\pi - |x||$ is an even function, which means the integral over the symmetric interval $[-\pi, \pi]$ is equal to $2\int_{0}^{\pi} (\pi - |x|) dx$. Inside this range, $|x|$ simplifies to $x$, and because $x \leq \pi$, the expression $(\pi - x)$ remains non-negative, allowing us to calculate $2\int_{0}^{\pi} (\pi - x) dx$ directly. By finding the antiderivative $2(\pi x - \frac{x^{2}}{2})$ and evaluating it at the boundaries $0$ and $\pi$, we obtain $2(\pi^{2} - \frac{\pi^{2}}{2})$. This calculation results in a final value of $\pi^{2}$.

The integral $\int_{0}^{3} (g(x) - f(x)) \, dx$ represents the net difference between the areas bounded by the curves $g(x)$ and $f(x)$ on the interval $[0, 3]$. Analyzing $f(x) = |x-2|$, the graph forms two distinct triangular regions with areas of 2 and 0.5, resulting in a total area of 2.5 under the curve. For $g(x) = f(f(x))$, which equals $||x-2|-2|$, the composite function creates a transformed geometric area defined by the sum of $(\frac{1}{2} \times 2 \times 2 + 1 + \frac{1}{2} \times 1 \times 1)$, which totals 3.5. By subtracting the total area of $f(x)$ from $g(x)$, the calculation $3.5 - 2.5$ yields the final result of 1.

The integrand is the result of applying the product rule for differentiation to the function $f(x) = \frac{1}{2} \sin^4 3x \tan^4 x$. Expanding the original expression by substituting $\sin 6x = 2 \sin 3x \cos 3x$ transforms the integrand into $2 \tan^3 x \sec^2 x \sin^4 3x + 6 \sin^3 3x \cos 3x \tan^4 x$, which is exactly $\frac{d}{dx} (\frac{1}{2} \sin^4 3x \tan^4 x)$. The definite integral is evaluated as $[\frac{1}{2} \sin^4 3x \tan^4 x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$. At the upper limit $x = \frac{\pi}{3}$, $\sin 3x = \sin \pi = 0$, making the term $0$. At the lower limit $x = \frac{\pi}{6}$, $\sin 3x = \sin \frac{\pi}{2} = 1$ and $\tan x = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$, resulting in the value $\frac{1}{2} [ 0 - (1)^4 (\frac{1}{\sqrt{3}})^4 ]$. Simplifying this expression leads to $-\frac{1}{18}$.

Evaluating the integral of a nested absolute value function requires partitioning the domain into intervals based on where the internal expressions change sign. For $x \ge 1$, the term $|x-1|-x$ simplifies to $(x-1)-x = -1$, resulting in a constant function value of $|-1| = 1$. When $0 \le x < 1$, the inner term $|x-1|$ becomes $1-x$, making the expression $|(1-x)-x| = |1-2x|$, which is further defined as $1-2x$ for $0 \le x < \frac{1}{2}$ and $2x-1$ for $\frac{1}{2} \le x < 1$. Summing the definite integrals over these three intervals, specifically $\int_{0}^{\frac{1}{2}} (1-2x) \, dx + \int_{\frac{1}{2}}^{1} (2x-1) \, dx + \int_{1}^{2} 1 \, dx$, yields $0.25 + 0.25 + 1$, which totals 1.5.

The absolute value function requires splitting the integral at $x = 0$, where the expression $|x|$ changes behavior from $-x$ to $x$. This leads to the following setup:

$4\alpha \left( \int_{-1}^{0} e^{\alpha x} dx + \int_{0}^{2} e^{-\alpha x} dx \right) = 5$

Executing the integration provides:

$4\alpha \left( \left[ \frac{e^{\alpha x}}{\alpha} \right]_{-1}^{0} + \left[ \frac{e^{-\alpha x}}{-\alpha} \right]_{0}^{2} \right) = 5$

Simplifying this expression results in $4(1 - e^{-\alpha} + 1 - e^{-2\alpha}) = 5$, which arranges into the quadratic equation $4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$. By substituting $t = e^{-\alpha}$, we solve the quadratic $4t^2 + 4t - 3 = 0$ to find the roots $1/2$ and $-3/2$. Since $e^{-\alpha}$ must be positive, we accept only $t = 1/2$, which leads to $e^{-\alpha} = 1/2$, or $\alpha = \log_e 2$.

Substituting $x=0$ directly into the expression reveals an indeterminate $\frac{0}{0}$ form, signaling the need for L'Hôpital's Rule to resolve the limit. Differentiating the numerator with respect to $x$ using the Fundamental Theorem of Calculus yields $x\sin(10x)$, while differentiating the denominator gives 1. Consequently, the limit simplifies to $\lim_{x\to 0} \frac{x\sin(10x)}{1}$, which evaluates to $0 \cdot \sin(0) = 0$.