Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

When the difference between function values is bounded by $|x - y|^{3/2}$, dividing both sides of the inequality by $|x - y|$ reveals that the magnitude of the difference quotient $\left|\frac{f(x) - f(y)}{x - y}\right|$ is bounded by $|x - y|^{1/2}$. As $x$ approaches $y$, the right side of this inequality tends toward 0, which forces the derivative $f'(y)$ to be identically zero for all values. Because the derivative is zero, the function $f(x)$ must be constant, and applying the condition $f(0) = 1$ confirms that $f(x) = 1$ for all $x$. Therefore, the definite integral $\int_{0}^{1} f^2(x) dx$ becomes $\int_{0}^{1} 1^2 dx$, which yields a total value of 1.

Evaluating integrals involving the greatest integer function relies on partitioning the interval into sub-intervals where the values of the floor expressions remain constant. Given the range from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, we split the integral based on the behavior of $\lfloor x \rfloor$ and $\lfloor \sin x \rfloor$.

In the interval $[-\frac{\pi}{2}, -1)$, $\lfloor x \rfloor = -2$ and $\lfloor \sin x \rfloor = -1$, simplifying the denominator to $-2 - 1 + 4 = 1$. The integral contribution is:
$\int_{-\frac{\pi}{2}}^{-1} dx = \frac{\pi}{2} - 1$
For the interval $[-1, 0)$, $\lfloor x \rfloor = -1$ and $\lfloor \sin x \rfloor = -1$, resulting in a denominator of $-1 - 1 + 4 = 2$. The integral contribution is:
$\int_{-1}^{0} \frac{1}{2} dx = \frac{1}{2}$
In the positive range $[0, 1)$, both $\lfloor x \rfloor$ and $\lfloor \sin x \rfloor$ equal $0$, making the denominator $0 + 0 + 4 = 4$. The integral contribution is:
$\int_{0}^{1} \frac{1}{4} dx = \frac{1}{4}$
Finally, for the interval $[1, \frac{\pi}{2}]$, $\lfloor x \rfloor = 1$ and $\lfloor \sin x \rfloor = 0$, leading to a denominator of $1 + 0 + 4 = 5$. The integral contribution is:
$\int_{1}^{\frac{\pi}{2}} \frac{1}{5} dx = \frac{1}{5} \left( \frac{\pi}{2} - 1 \right) = \frac{\pi}{10} - \frac{1}{5}$

Summing these individual parts provides the total value:
$\left( \frac{\pi}{2} - 1 \right) + \frac{1}{2} + \frac{1}{4} + \left( \frac{\pi}{10} - \frac{1}{5} \right) = \frac{\pi}{2} + \frac{\pi}{10} - 1 + \frac{1}{2} + \frac{1}{4} - \frac{1}{5}$
Combining the terms gives:
$\frac{6\pi}{10} + \frac{-20 + 10 + 5 - 4}{20} = \frac{6\pi}{10} - \frac{9}{20}$
This expression simplifies to:
$\frac{12\pi - 9}{20} = \frac{3}{20} (4\pi - 3)$

To minimize the definite integral of a function, we must integrate over the entire region where the function takes negative values, as including any portion where the function is positive would increase the result. Analyzing the integrand $f(x) = x^4 - 2x^2$, we can rewrite it by factoring as
$f(x) = x^2(x^2 - 2)$
This expression identifies the roots at $x = 0$ and $x = \pm \sqrt{2}$. Because $x^2$ is always non-negative, the function values are negative only when $x^2 - 2 < 0$, which corresponds to the interval $(-\sqrt{2}, \sqrt{2})$. Integrating across this entire range captures all negative values, ensuring that the integral reaches its absolute minimum.

The King's property of definite integrals, which states that $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$, is the primary tool for simplifying expressions involving the greatest integer function. Applying this transformation over the interval from $0$ to $2\pi$ leads to $\sin(2(2\pi - x)) = -\sin 2x$ and $\cos(3(2\pi - x)) = \cos 3x$, which converts the integrand $f(x)$ into $-f(x)$. Integrating the sum of the original and the transformed function yields $\int_0^{2\pi} -1 \, dx$, as the identity $[y] + [-y] = -1$ holds almost everywhere. Computing this integral gives $-2\pi$, and since this sum represents double the original value, the final result is $-\pi$.

To find the limit of $g(x)$ as $x$ approaches 2, we rearrange the given equation to isolate $g(x)$ as $g(x) = \frac{\int_{6}^{f(x)} 4 t^{3} \, dt}{x-2}$. This expression results in an indeterminate form, so we apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$. According to the Fundamental Theorem of Calculus, the derivative of the numerator is $4(f(x))^3 \cdot f'(x)$, while the derivative of the denominator is 1. Substituting the value $x=2$ and recognizing that the integral vanishes at this point when $f(2)=6$, we calculate the limit as $4(6)^3 \cdot \frac{1}{48}$, which simplifies to 18.

Simplifying the integrand by expressing all terms in powers of $\tan x$ provides a clear path to the result. Given that $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$ and $\cot^5 x = \frac{1}{\tan^5 x}$, the integral transforms into the form
$\frac{1}{2} \int \frac{\tan^4 x \sec^2 x}{1 + \tan^{10} x} \, dx$
Substituting $u = \tan^5 x$ allows the numerator to be expressed as $\frac{1}{5} \, du$, leading to the integral $\frac{1}{10} \int \frac{du}{1 + u^2}$. As $x$ ranges from $\frac{\pi}{6}$ to $\frac{\pi}{4}$, the corresponding values for $u$ are $(\frac{1}{\sqrt{3}})^5$ and $1^5$, respectively. Since $(\frac{1}{\sqrt{3}})^5 = \frac{1}{9\sqrt{3}}$, the expression is evaluated as
$\frac{1}{10} \left[ \tan^{-1} u \right]_{\frac{1}{9\sqrt{3}}}^{1}$
This calculation yields the final result of $\frac{1}{10} \left( \frac{\pi}{4} - \tan^{-1}\left( \frac{1}{9\sqrt{3}} \right) \right)$.

The substitution $x^2 = t$ simplifies the integrand, reducing the problem to $\frac{1}{2} \int_{0}^{1} \cot^{-1}(1-t+t^2) dt$. Utilizing the identity $\cot^{-1}(y) = \tan^{-1}(1/y)$, the argument can be decomposed into $\tan^{-1} t + \tan^{-1}(1-t)$, which follows from the relation $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$. Since the integrals of $\tan^{-1} t$ and $\tan^{-1}(1-t)$ are identical over the interval from 0 to 1, the expression simplifies further to $\int_{0}^{1} \tan^{-1} t dt$. Integrating by parts gives $t \tan^{-1} t - \frac{1}{2} \log_e(1+t^2)$, and evaluating this from 0 to 1 leads directly to $\frac{\pi}{4} - \frac{1}{2} \log_e 2$.

The integral simplifies by recognizing that $\frac{\tan\theta}{\sqrt{\sec\theta}}$ is equivalent to $\frac{\sin\theta}{\sqrt{\cos\theta}}$, allowing the entire expression to be rewritten as $\frac{1}{\sqrt{2k}} \int_{0}^{\pi/3} \sin\theta(\cos\theta)^{-1/2} \, d\theta$. Evaluating the integral of $\sin\theta(\cos\theta)^{-1/2}$ yields $-2\sqrt{\cos\theta}$, which, when applied over the limits $0$ to $\pi/3$, results in $-2\left( \sqrt{\cos(\pi/3)} - \sqrt{\cos(0)} \right)$. This simplifies to $-2\left( \frac{1}{\sqrt{2}} - 1 \right)$, which is equal to $2\left( 1 - \frac{1}{\sqrt{2}} \right)$. By setting $\frac{1}{\sqrt{2k}} \cdot 2\left( 1 - \frac{1}{\sqrt{2}} \right)$ equal to the given value of $1 - \frac{1}{\sqrt{2}}$, we determine that $\frac{2}{\sqrt{2k}} = 1$, which leads to $\sqrt{2k} = 2$ and confirms that $k = 2$.

To evaluate an integral over a symmetric interval like $[-2, 2]$, we can effectively pair the function's values at $x$ and $-x$ to simplify the integrand. We represent the integral as $\int_{0}^{2} (f(x) + f(-x)) dx$, where $f(x) = \frac{\sin^{2} x}{\lfloor x/\pi \rfloor + 1/2}$. The combined integrand becomes:

$\frac{\sin^{2} x}{\lfloor x/\pi \rfloor + 1/2} + \frac{\sin^{2} (-x)}{\lfloor -x/\pi \rfloor + 1/2}$

Given that $\sin^{2}(-x) = \sin^{2} x$, the numerators are identical. For non-integer values of $u$, the property $\lfloor u \rfloor + \lfloor -u \rfloor = -1$ holds, which allows us to rewrite the denominator of the second term. Specifically, because $\lfloor -x/\pi \rfloor = -1 - \lfloor x/\pi \rfloor$, the denominator of the second term becomes $-1 - \lfloor x/\pi \rfloor + 1/2$, which simplifies to $-(\lfloor x/\pi \rfloor + 1/2)$. Since the two fractions are additive opposites, they sum to zero, leading to an overall integral value of zero.

The integrand $\frac{\cot x}{\cot x + \csc x}$ simplifies efficiently by expressing the trigonometric terms in their sine and cosine forms. Multiplying the numerator and denominator by $\sin x$ yields $\frac{\cos x}{\cos x + 1}$, which can be rearranged as $1 - \frac{1}{1+\cos x}$. By applying the identity $1 + \cos x = 2\cos^2(\frac{x}{2})$, the integral becomes $\int_{0}^{\frac{\pi}{2}} (1 - \frac{1}{2}\sec^2(\frac{x}{2})) \, dx$.

Evaluating this integral leads to $[x - \tan(\frac{x}{2})]_{0}^{\frac{\pi}{2}}$. Substituting the upper and lower bounds results in $(\frac{\pi}{2} - \tan(\frac{\pi}{4})) - (0 - \tan(0))$, which simplifies to $\frac{\pi}{2} - 1$. Factoring out $\frac{1}{2}$ brings the expression to the form $\frac{1}{2}(\pi - 2)$. Matching this to $m(\pi + n)$ reveals $m = \frac{1}{2}$ and $n = -2$, meaning the product $m \cdot n$ equals $-1$.

The core principle here is to simplify the complex integrand by splitting the expression into two separate parts and employing integration by substitution for each. This technique transforms the complicated exponential and logarithmic functions into basic, manageable forms.

Splitting the integral, we evaluate $\int_{1}^{e} (\frac{x}{e})^{2x} \ln x \, dx - \int_{1}^{e} (\frac{e}{x})^x \ln x \, dx$. For the first term, substitute $t = (\frac{x}{e})^{2x}$. Taking the natural logarithm gives $\ln t = 2x(\ln x - 1)$, and differentiating yields $\frac{dt}{t} = 2 \ln x \, dx$, or $\ln x \, dx = \frac{dt}{2}$. As $x$ ranges from $1$ to $e$, $t$ ranges from $(1/e)^2$ to $1$. The first integral becomes $\int_{(1/e)^2}^{1} \frac{1}{2} dt = \frac{1}{2}(1 - \frac{1}{e^2})$

For the second term, substitute $v = (\frac{e}{x})^x$. Logarithmic differentiation gives $\frac{dv}{v} = -\ln x \, dx$, which implies $\ln x \, dx = -\frac{dv}{v}$. Thus, the term $-\int_{1}^{e} (\frac{e}{x})^x \ln x \, dx$ simplifies to $-\int_{1}^{e} v (-\frac{dv}{v}) = \int_{1}^{e} dv$. Since $v(1) = e$ and $v(e) = 1$, the integral evaluates to $v|_{1}^{e} = 1 - e$

Combining these, the total value is $\frac{1}{2}(1 - \frac{1}{e^2}) + (1 - e) = \frac{3}{2} - e - \frac{1}{2e^2}$

Decomposing the integrand $\frac{1}{(x+\alpha)(x+\alpha+1)}$ into partial fractions yields $\frac{1}{x+\alpha} - \frac{1}{x+\alpha+1}$, which integrates to $\ln \left| \frac{x+\alpha}{x+\alpha+1} \right|$. Evaluating this expression over the limits from $\alpha$ to $\alpha+1$ produces $\ln \left| \frac{2\alpha+1}{2\alpha+2} \right| - \ln \left| \frac{2\alpha}{2\alpha+1} \right|$, which simplifies to $\ln \left| \frac{(2\alpha+1)^2}{4\alpha(\alpha+1)} \right|$. Equating this result to $\ln \left( \frac{9}{8} \right)$ creates the algebraic relationship $\frac{(2\alpha+1)^2}{4\alpha(\alpha+1)} = \frac{9}{8}$, which rearranges to $8(4\alpha^2 + 4\alpha + 1) = 36\alpha^2 + 36\alpha$. Solving the resulting quadratic equation $4\alpha^2 + 4\alpha - 8 = 0$, or $\alpha^2 + \alpha - 2 = 0$, confirms that $\alpha = -2$ satisfies the condition.

L'Hôpital's Rule applies because substituting $x=2$ directly into the expression results in the indeterminate form $\frac{0}{0}$, given that $f(2)=6$ makes the integral $\int_{6}^{6} 2t\,dt$ equal to $0$. We differentiate the numerator using the Fundamental Theorem of Calculus combined with the chain rule:

$\frac{d}{dx} \int_{6}^{f(x)} 2t\,dt = 2f(x) \cdot f'(x)$

The derivative of the denominator $x-2$ with respect to $x$ is $1$. Applying L'Hôpital's Rule, the original limit simplifies to $\lim_{x\to 2} 2 f(x) f'(x)$, and substituting $f(2)=6$ into this expression yields $2 \cdot 6 \cdot f'(2)$, which equals $12 f'(2)$.

Leibniz's rule for differentiation under the integral sign provides the fundamental approach for solving equations containing variable integral limits. Differentiating both sides of the equation with respect to $x$ yields $f(x)$ on the left, while the right side becomes $2x - x^2 f(x)$ since the derivative of the integral $\int_{x}^{1} t^2 f(t) dt$ results in $-x^2 f(x)$.

Rearranging this relationship gives $f(x)(1 + x^2) = 2x$, which allows us to define the function as $f(x) = \frac{2x}{1+x^2}$. To find the required value, we apply the quotient rule to calculate the derivative, resulting in $f'(x) = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2}$, which simplifies to $\frac{2-2x^2}{(1+x^2)^2}$. Substituting $x = \frac{1}{2}$ into this derivative leads to $\frac{2-2(1/4)}{(1+1/4)^2}$, yielding $\frac{3/2}{25/16}$, which simplifies to $\frac{24}{25}$.

The property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx$ allows for the evaluation of integrals over symmetric intervals by analyzing the parity of the integrand. The integral can be decomposed into two distinct parts: $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \, dx$ and $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4 x \log\left(\frac{2+\sin x}{2-\sin x}\right) \, dx$. Defining the integrand of the second part as $g(x) = \sin^4 x \log\left(\frac{2+\sin x}{2-\sin x}\right)$, one observes that $g(-x) = \sin^4(-x) \log\left(\frac{2-\sin x}{2+\sin x}\right) = \sin^4 x \left(-\log\left(\frac{2+\sin x}{2-\sin x}\right)\right) = -g(x)$. Since $g(x)$ is an odd function, its integral over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is zero. Consequently, the integral simplifies to $2 \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx$. Applying Wallis's formula, where $\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{(n-1)!!}{n!!} \frac{\pi}{2}$ for an even integer $n$, the value is calculated as $2 \left( \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} \right) = \frac{3\pi}{8}$.

Definite integrals over symmetric intervals are often resolved using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$. Applying this to the integral $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^x} \, dx$, where the bounds sum to zero, allows for the transformation $x \to -x$. Substituting $-x$ for $x$ yields $I = \int_{-\pi/2}^{\pi/2} \frac{\sin^2(-x)}{1+2^{-x}} \, dx$, which simplifies using $\sin^2(-x) = \sin^2 x$ and $2^{-x} = 1/2^x$ to produce $I = \int_{-\pi/2}^{\pi/2} \frac{2^x \sin^2 x}{1+2^x} \, dx$.

Combining the original expression for $I$ and this transformed version eliminates the exponential term from the denominator:
$2I = \int_{-\pi/2}^{\pi/2} \left( \frac{\sin^2 x}{1+2^x} + \frac{2^x \sin^2 x}{1+2^x} \right) \, dx = \int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$
Because $\sin^2 x$ is an even function, the integral can be written as $2 \int_{0}^{\pi/2} \sin^2 x \, dx$. The definite integral $\int_{0}^{\pi/2} \sin^2 x \, dx$ equals $\pi/4$, resulting in $2I = 2 \times (\pi/4) = \pi/2$, which simplifies to $I = \pi/4$.

The function is defined as $f(x) = \sin x \int_0^x t \, dt - \int_0^x t \sin t \, dt$. Evaluating the integrals, we get $f(x) = \sin x \left( \frac{x^2}{2} \right) - \left( [-t \cos t]_0^x + \int_0^x \cos t \, dt \right) = \frac{x^2}{2} \sin x + x \cos x - \sin x$. Differentiating this expression with respect to $x$ gives $f'(x) = \frac{d}{dx} \left( \frac{x^2}{2} \sin x + x \cos x - \sin x \right) = x \sin x + \frac{x^2}{2} \cos x + \cos x - x \sin x - \cos x = \frac{x^2}{2} \cos x$. Further differentiating to find the higher-order derivatives, $f''(x) = \frac{d}{dx} \left( \frac{x^2}{2} \cos x \right) = x \cos x - \frac{x^2}{2} \sin x$, and $f'''(x) = \frac{d}{dx} \left( x \cos x - \frac{x^2}{2} \sin x \right) = \cos x - x \sin x - x \sin x - \frac{x^2}{2} \cos x = \cos x - 2x \sin x - \frac{x^2}{2} \cos x$. Combining these results, $f'''(x) + f'(x) = \left( \cos x - 2x \sin x - \frac{x^2}{2} \cos x \right) + \frac{x^2}{2} \cos x = \cos x - 2x \sin x$.

The comparison property of definite integrals states that if two functions satisfy $f(x) \le g(x)$ for all $x \in [a, b]$, then $\int_a^b f(x) dx \le \int_a^b g(x) dx$. In the interval $x \in (0, 1)$, the relationship $x > x^2 > x^3$ holds, which implies $-x < -x^2 < -x^3$, and leads to the exponential inequality $e^{-x} < e^{-x^2} < e^{-x^3}$. Given that $\cos^2 x \le 1$ and $\cos^2 x > 0$ on this interval, multiplying the terms results in the integrand inequality $e^{-x} \cos^2 x < e^{-x^2} \cos^2 x \le e^{-x^2} \le e^{-x^3}$. Integrating each expression from $0$ to $1$ preserves this order, yielding $I_1 < I_2 < I_3$, which is equivalent to $I_3 > I_2 > I_1$.

The property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$ simplifies the evaluation of definite integrals by utilizing symmetry. Applying this to $I = \int_{\pi/4}^{3\pi/4} \frac{x}{1+\sin x} \, dx$, where the sum of limits $a+b$ equals $\pi$, substitutes $x$ with $\pi-x$ to produce $I = \int_{\pi/4}^{3\pi/4} \frac{\pi-x}{1+\sin x} \, dx$. Summing these two identical expressions for $I$ results in $2I = \int_{\pi/4}^{3\pi/4} \frac{\pi}{1+\sin x} \, dx$, which simplifies to $I = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \frac{1}{1+\sin x} \, dx$.

Multiplying the integrand by the conjugate $\frac{1-\sin x}{1-\sin x}$ transforms the expression into $\frac{1-\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x$. Integrating this yields $I = \frac{\pi}{2} [\tan x - \sec x]_{\pi/4}^{3\pi/4}$. Evaluating at the limits gives $I = \frac{\pi}{2} [(\tan(3\pi/4) - \sec(3\pi/4)) - (\tan(\pi/4) - \sec(\pi/4))] = \frac{\pi}{2} [(-1 - (-\sqrt{2})) - (1 - \sqrt{2})] = \frac{\pi}{2} [-1+\sqrt{2}-1+\sqrt{2}] = \frac{\pi}{2} [2\sqrt{2}-2]$, which simplifies to $\pi(\sqrt{2}-1)$.

The evaluation of this definite integral is best approached by utilizing the symmetry property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. Since the sum of the limits $\frac{\pi}{4} + \frac{3\pi}{4}$ equals $\pi$, replacing $x$ with $\pi-x$ transforms the integrand $\frac{1}{1+\cos x}$ into $\frac{1}{1-\cos x}$, allowing us to set the integral $I$ equal to itself in this new form and combine them as follows:

$2I = \int_{\pi/4}^{3\pi/4} \left( \frac{1}{1+\cos x} + \frac{1}{1-\cos x} \right) \, dx = \int_{\pi/4}^{3\pi/4} \frac{2}{\sin^2 x} \, dx$

This expression simplifies to $I = \int_{\pi/4}^{3\pi/4} \csc^2 x \, dx$. Evaluating the antiderivative $-\cot x$ between the limits $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ results in $-\cot(\frac{3\pi}{4}) - (-\cot(\frac{\pi}{4}))$, which becomes $-(-1) - (-1)$, confirming the final value is 2.