Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Evaluating local extrema for a function defined by an integral, such as $f(x) = \int_{0}^{x^2} \frac{t^{2-}5t+4}{2+e^t} \, dt$, begins by applying the Leibniz rule to differentiate the function. This process results in $f'(x) = \frac{(x^2)^2 - 5(x^2) + 4}{2+e^{x^2}} \cdot (2x)$, which simplifies to the expression $\frac{2x(x-1)(x+1)(x-2)(x+2)}{2+e^{x^2}}$. Setting this derivative equal to zero identifies the critical points at $x = 0, \pm 1, \pm 2$.

Since the denominator $2+e^{x^2}$ remains strictly positive for all real values of $x$, the sign of the derivative is determined entirely by the polynomial in the numerator. Examining the sign changes across the critical points reveals that $f'(x)$ switches from negative to positive at $x = -2$, $x = 0$, and $x = 2$, identifying these as local minimum points. Conversely, the derivative switches from positive to negative at $x = -1$ and $x = 1$, identifying these as local maximum points. Based on this analysis, the function possesses two local maximum points and three local minimum points, meaning $m=2$ and $n=3$.

Applying the Leibniz rule for differentiation under the integral sign to the integral $\int_{\cos x}^{1} t^2 f(t) \, dt = \sin^3 x + \cos x$ allows us to extract information about the function $f$. Differentiating both sides with respect to $x$ results in the left side becoming $\sin x \cos^2 x f(\cos x)$, as the derivative of the integral with respect to its lower limit $\cos x$ contributes a term of $- (\cos x)^2 f(\cos x) \cdot \frac{d}{dx}(\cos x)$, which simplifies to $\sin x \cos^2 x f(\cos x)$. The derivative of the right side is $3 \sin^2 x \cos x - \sin x$. Equating these expressions and dividing by $\sin x$, which is non-zero in the given interval, reveals that $\cos^2 x f(\cos x) = 3 \sin x \cos x - 1$.

Dividing further by $\cos^2 x$ leads to the functional form $f(\cos x) = 3 \tan x - \sec^2 x$. Differentiating this equation with respect to $x$ requires the chain rule on the left, yielding $f'(\cos x) \cdot (-\sin x)$, while the right side differentiates to $3 \sec^2 x - 2 \sec^2 x \tan x$. At the point where $\cos x = \frac{1}{\sqrt{3}}$, we have $\sin x = \frac{\sqrt{2}}{\sqrt{3}}$, $\tan x = \sqrt{2}$, and $\sec^2 x = 3$. Substituting these values into the differentiated equation gives $-\frac{\sqrt{2}}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = 3(3) - 2(3)(\sqrt{2})$, which simplifies to $-\frac{\sqrt{2}}{\sqrt{3}} f'\left(\frac{1}{\sqrt{3}}\right) = 9 - 6\sqrt{2}$. Solving for $f'\left(\frac{1}{\sqrt{3}}\right)$ results in $f'\left(\frac{1}{\sqrt{3}}\right) = -\frac{\sqrt{3}}{\sqrt{2}}(9 - 6\sqrt{2}) = 6\sqrt{3} - \frac{9\sqrt{3}}{\sqrt{2}}$. Dividing this by $\sqrt{3}$ gives the final value of $6 - \frac{9}{\sqrt{2}}$.

The limit expression $a$ represents a Riemann sum, which can be identified by rewriting the sum as $a = \lim_{n \to \infty } \sum_{k=1}^{n} \frac{1}{n} \frac{2}{1+(k/n)^2}$. This structure corresponds to the definite integral
$a = \int_{0}^{1} \frac{2}{1+x^{2}} \, dx = 2 \tan^{-1}(1) = \frac{\pi}{2}$
For the function $f(x) = \sqrt{\frac{1-\cos x}{1+\cos x}}$, applying trigonometric half-angle identities simplifies the expression to $f(x) = \tan(x/2)$. Differentiating this function with respect to $x$ yields $f'(x) = \frac{1}{2} \sec^2(x/2)$. Evaluating these expressions at $x = a/2 = \pi/4$, we find $f(\pi/4) = \tan(\pi/8) = \sqrt{2}-1$ and $f'(\pi/4) = \frac{1}{2} \sec^2(\pi/8)$. Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$, we calculate the derivative as $f'(\pi/4) = \frac{1}{2}(1 + (\sqrt{2}-1)^2) = 2 - \sqrt{2}$. Since multiplying the value of $f(\pi/4)$ by $\sqrt{2}$ yields $\sqrt{2}(\sqrt{2}-1) = 2 - \sqrt{2}$, the equality $\sqrt{2}f(a/2) = f'(a/2)$ is confirmed.

The function $f(x) = \max\{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$ represents the upper envelope of these individual absolute value expressions over the interval $[-6, 0]$. Comparing the magnitudes reveals that the expression $|x+1|$ dominates from $x=-6$ to $x=-3$, whereas $|x+5|$ represents the maximum value from $x=-3$ to $x=0$. These two linear components intersect where $-x-1 = x+5$, which occurs exactly at $x=-3$.

To find the integral $\int_{-6}^{0} f(x) \, dx$, we split the calculation into two parts at the point of intersection. The first integral, $\int_{-6}^{-3} (-x-1) \, dx$, evaluates to $[-\frac{x^2}{2} - x]_{-6}^{-3}$, which simplifies to $(-\frac{9}{2} + 3) - (-18 + 6) = -1.5 - (-12) = 10.5$. The second integral, $\int_{-3}^{0} (x+5) \, dx$, evaluates to $[\frac{x^2}{2} + 5x]_{-3}^{0}$, resulting in $0 - (\frac{9}{2} - 15) = -(-10.5) = 10.5$. Summing these two areas gives a total result of $21$.

Integrals with reciprocal limits often rely on symmetry to simplify complex terms. Considering the sum $f(\alpha) + f(1/\alpha)$, the substitution $t = 1/x$ in the integral for $f(1/\alpha)$ transforms the limits from $1$ and $1/\alpha$ to $1$ and $\alpha$, while the differential $dt$ becomes $-\frac{1}{x^2} \, dx$. The integrand simplifies to $\frac{\log_{10} x}{x(x+1)}$, which decomposes into $\frac{\log_{10} x}{x} - \frac{\log_{10} x}{x+1}$. Adding this to $f(\alpha)$ causes the term $\frac{\log_{10} t}{1+t}$ to cancel out, leaving the integral $\int_{1}^{\alpha} \frac{\log_{10} t}{t} \, dt$.

Converting the logarithm to natural base gives $\frac{1}{\ln 10} \int_{1}^{\alpha} \frac{\ln t}{t} \, dt$. Since $\frac{1}{t} \, dt$ is the derivative of $\ln t$, evaluating the integral results in $\frac{(\ln \alpha)^2}{2 \ln 10}$. Substituting $\alpha = e^{3}$ into this expression yields $\frac{(\ln e^{3})^2}{2 \ln 10}$, which simplifies to $\frac{9}{2 \ln 10}$ or $\frac{9}{2 \log_{e} 10}$.

Limits of summations as $n$ approaches infinity are effectively interpreted as definite integrals by identifying a Riemann sum structure. By rewriting the summation as $\frac{1}{n} \sum_{r=1}^{n} \frac{r/n}{2(r/n)^2 - 7(r/n) + 6}$, each term containing $r/n$ is treated as $x$, and the scaling factor $\frac{1}{n}$ represents the differential $dx$. As $n$ tends toward infinity, the lower bound $r=1$ corresponds to $x=0$, and the upper bound $r=n$ corresponds to $x=1$, allowing the expression to be evaluated as $\int_{0}^{1} \frac{x}{2x^2 - 7x + 6} \, dx$.

The quadratic denominator factors into $(2x - 3)(x - 2)$, which enables the integrand to be decomposed into partial fractions of the form $\frac{-3}{2x - 3} + \frac{2}{x - 2}$. Integrating these terms individually yields $-\frac{3}{2} \ln|2x - 3| + 2 \ln|x - 2|$. Evaluating this antiderivative across the limits from $0$ to $1$, the value at the upper limit is zero. Subtracting the value at the lower limit results in $\frac{3}{2} \ln(3) - 2 \ln(2)$. Applying the laws of logarithms, this simplifies to $\ln\left(3^{3/2}\right) - \ln\left(2^2\right)$, which is $\ln\left(\frac{3\sqrt{3}}{4}\right)$.

The expression $\frac{\sin 6x}{\sin x}$ is most efficiently handled by expanding $\sin 6x$ as $2 \sin 3x \cos 3x$. Utilizing the triple angle identities, where $\sin 3x = \sin x (3 - 4 \sin^2 x)$ and $\cos 3x = \cos x (4 \cos^2 x - 3)$, the ratio simplifies into $2(3 - 4 \sin^2 x)(4 \cos^2 x - 3) \cos x$.

Multiplying this by the constant factor $60$ yields an integrand well-suited for substitution. By setting $t = \sin x$, the differential becomes $dt = \cos x dx$, and converting the $\cos^2 x$ term to $1 - t^2$ results in the factor $(4(1 - t^2) - 3)$, which simplifies to $1 - 4t^2$. As the limits of integration change from $[0, \frac{\pi}{2}]$ to $[0, 1]$, the integral transforms into $120 \int_{0}^{1} (3 - 4t^2)(1 - 4t^2) dt$.

Expanding the integrand gives $3 - 12t^2 - 4t^2 + 16t^4$, which simplifies to $3 - 16t^2 + 16t^4$. Integrating this polynomial term by term results in $120 [3t - \frac{16t^3}{3} + \frac{16t^5}{5}]$, evaluated from $0$ to $1$. Computing the final value produces $120(3 - \frac{16}{3} + \frac{16}{5})$, which simplifies to $120(\frac{13}{15})$, resulting in $104$.

The integral is solved by first partitioning the interval $[0, 1]$ at the roots of the quadratic expression $3x^{2} - 5x + 2 = 0$, which occur at $x = 2/3$ and $x = 1$. Within the absolute value, $3x^{2} - 5x + 2$ is non-negative on $[0, 2/3]$ and non-positive on $[2/3, 1]$. Applying the property $\lfloor u + 1 \rfloor = \lfloor u \rfloor + 1$ for any integer constant allows the integral to be expressed as $\int_{0}^{1} \lfloor 2x - |3x^{2} - 5x + 2| \rfloor \, dx + 1$.

For the first sub-interval $x \in [0, 2/3]$, the expression becomes $\lfloor 2x - (3x^{2} - 5x + 2) \rfloor = \lfloor -3x^{2} + 7x - 2 \rfloor$. The floor function changes value at the points where $-3x^{2} + 7x - 2$ equals an integer. Solving $-3x^{2} + 7x - 2 = k$ for $k = -2, -1, 0, 1$ yields the partition points $x = 0$, $\alpha = \frac{7 - \sqrt{37}}{6}$, $x = 1/3$, $\beta = \frac{7 - \sqrt{13}}{6}$, and $x = 2/3$. The integral over this interval evaluates to $-2\alpha - 1(1/3 - \alpha) + 0(\beta - 1/3) + 1(2/3 - \beta) = -\alpha - \beta + 1/3$.

For the second sub-interval $x \in [2/3, 1]$, the expression inside the floor is $2x - (-(3x^{2} - 5x + 2)) = 3x^{2} - 3x + 2$. On this range, the function $3x^{2} - 3x + 2$ takes values in $[4/3, 2]$, so its floor is $1$. The integral for this part is therefore $\int_{2/3}^{1} 1 \, dx = 1/3$. Combining these results with the earlier constant $1$, the total integral is $(-\alpha - \beta + 1/3) + 1/3 + 1 = 5/3 - (\alpha + \beta)$. Substituting the values $\alpha = \frac{7 - \sqrt{37}}{6}$ and $\beta = \frac{7 - \sqrt{13}}{6}$, the final expression becomes $\frac{\sqrt{37} + \sqrt{13} - 4}{6}$.

Evaluating the definite integral $a_n = \int_{-1}^{n} \left(1 + \frac{x}{2} + \frac{x^{2}}{3} + \dots + \frac{x^{n-1}}{n}\right) dx$ requires determining the function's value for increasing integer values of $n$ to identify which ones fall within the specified range $(2, 30)$. By integrating the power series term-by-term from $-1$ to $n$, we obtain a sequence of values for $a_n$. For $n=1$, the integral evaluates to $2$. When $n=2$, the calculation yields $4.5$, and for $n=3$, the value is $\frac{100}{9}$, which is approximately $11.11$. Continuing to $n=4$, the integral results in a value greater than $31$, placing it outside the required interval. Since the condition $a_n \in (2, 30)$ is satisfied only by $n=2$ and $n=3$, the sum of these valid elements is $5$.

Approximating the limit of an infinite sum as a definite integral transforms complex summations into a manageable area calculation. The series $\sum_{r=1}^n \frac{n^2}{(n^{2+}r^2)(n+r)}$ can be rewritten by factoring out $n^3$ to obtain $\sum_{r=1}^n \frac{1}{n} \cdot \frac{1}{(1+(r/n)^2)(1+(r/n))}$, which represents the Riemann sum of $\int_0^1 \frac{1}{(1+x^2)(1+x)} \, dx$ as $n$ approaches infinity. Decomposing the integrand into $\frac{1}{2} \left( \frac{1}{1+x} - \frac{x-1}{1+x^2} \right)$ allows for straightforward term-by-term integration. Evaluating these components over the interval $[0, 1]$ yields $\frac{1}{2} \ln 2 - \frac{1}{4} \ln 2 + \frac{1}{2} \left( \frac{\pi}{4} \right)$, which simplifies to $\frac{\pi}{8} + \frac{1}{4} \ln 2$.

The core strategy involves applying a trigonometric substitution to simplify the complex radical in the denominator. By letting $x = \tan \theta$, the differential becomes $dx = \sec^2 \theta d\theta$, and the denominator simplifies using the identity $1 + \tan^2 \theta = \sec^2 \theta$. Specifically, the term under the square root transforms into $\sqrt{\sec^2 \theta + \sec^3 \theta}$, which simplifies further to $\sec \theta \sqrt{1 + \sec \theta}$. Consequently, the entire integral converts to $15 \int_{0}^{\pi/3} \frac{(\sec^2 \theta - 1) \sec \theta \tan \theta}{\sqrt{1 + \sec \theta}} d\theta$.

To resolve the integral, we perform a second substitution by setting $1 + \sec \theta = t^2$, which yields $\sec \theta \tan \theta d\theta = 2t dt$ and changes the integration limits from $0$ to $\pi/3$ into $\sqrt{2}$ to $\sqrt{3}$. This substitution reduces the integral to $30 \int_{\sqrt{2}}^{\sqrt{3}} (t^4 - 2t^2) dt$, which is straightforward to evaluate. Computing the antiderivative $30 \left( \frac{t^5}{5} - \frac{2t^3}{3} \right)$ at the given boundaries results in $16\sqrt{2} - 6\sqrt{3}$. By identifying the coefficients corresponding to the required form, we find $\alpha = 16$ and $\beta = -6$, resulting in a sum of $10$.

Definite integrals represent the area under a curve, allowing us to reconcile complex expressions by evaluating each side of an equation independently. Calculating the integral on the left side, $\int_{0}^{2} (\sqrt{2x} - \sqrt{2x-x^2}) \, dx$, yields a total value of $\frac{8}{3} - \frac{\pi}{2}$. When the integrals on the right side are calculated and combined, they simplify to the expression $I + \frac{5}{3} - \frac{\pi}{4}$. Equating these two results requires that $I$ be equal to $1 - \frac{\pi}{4}$, which is precisely the value produced by evaluating the integral $\int_{0}^{1} (1 - \sqrt{1-y^2}) \, dy$.

We first define the function $f(x)=2+|x|-|x-1|+|x+1|$ by considering the critical points at $x = -1, 0, 1$ to remove the absolute value bars. This produces a piecewise linear function where $f(x)=-x$ for $x < -1$, $f(x)=x+2$ for $-1 \leq x < 0$, $f(x)=3x+2$ for $0 \leq x < 1$, and $f(x)=x+4$ for $x \geq 1$.

Calculating the derivative at each requested point yields $f'(-\frac{3}{2})=-1$, $f'(-\frac{1}{2})=1$, $f'(\frac{1}{2})=3$, and $f'(\frac{3}{2})=1$. Summing these values gives $-1 + 1 + 3 + 1 = 4$, which contradicts the first statement claiming a sum of $2$.

To evaluate the second statement, we calculate the definite integral $\int_{-2}^{2} f(x) \, dx$ by partitioning it into four distinct intervals corresponding to the piecewise definition: $\int_{-2}^{-1} (-x) \, dx + \int_{-1}^{0} (x+2) \, dx + \int_{0}^{1} (3x+2) \, dx + \int_{1}^{2} (x+4) \, dx$. Computing the integral for each section results in $\frac{3}{2} + \frac{3}{2} + \frac{7}{2} + \frac{11}{2} = \frac{24}{2} = 12$. Since this matches the value in the second statement, we conclude that only the second statement is correct.

Calculus problems involving absolute value and greatest integer functions rely on splitting the interval of integration at points where the expression inside the absolute value changes sign or where the output of the greatest integer function shifts. The absolute value $|2x^2 - 3x|$ changes behavior at $x = 1.5$, requiring the integral to be separated into the intervals $[0, 1.5]$ and $[1.5, 2]$, where the expression functions as $-(2x^2 - 3x)$ and $(2x^2 - 3x)$ respectively, resulting in a calculated area of $\frac{19}{12}$. Simultaneously, the floor function $[x - 0.5]$ remains constant at $-1$ for $x \in [0, 0.5)$, $0$ for $x \in [0.5, 1.5)$, and $1$ for $x \in [1.5, 2]$, which sums to a net integral of zero across the entire range $[0, 2]$. Adding these components together yields a final value of $\frac{19}{12}$.

Evaluating this limit involves recognizing the indeterminate form, which permits the application of L'Hôpital's rule by differentiating the numerator and the denominator with respect to $x$. According to the Leibniz integral rule, differentiating the integral component in the numerator yields $\frac{\pi}{4} \cdot 2 \sec^2 x \cdot \tan x \cdot f(\sec^2 x)$, while the derivative of the denominator $x^2 - \frac{\pi^2}{16}$ is $2x$. Substituting $x = \frac{\pi}{4}$ into this expression relies on the trigonometric values $\sec^2(\frac{\pi}{4}) = 2$ and $\tan(\frac{\pi}{4}) = 1$. This leads to the simplified quotient $\frac{\frac{\pi}{4} \cdot 2 (2) (1) f(2)}{2 (\frac{\pi}{4})}$ which reduces to $2 f(2)$.

We can solve this problem by applying the definite integral property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. Substituting $x$ with $-x$ over the interval $[-\pi/2, \pi/2]$ transforms the integrand into $\frac{\pi^{\sin x}(1+\sin^2 x)}{1+\pi^{\sin x}}$, and adding this to the original integral $I$ effectively cancels out the denominator, leaving $2I = \int_{-\pi/2}^{\pi/2} (1+\sin^2 x) \, dx$. Since the integrand is an even function, we can simplify this to $2I = 2 \int_{0}^{\pi/2} (1+\sin^2 x) \, dx$, or $I = \int_{0}^{\pi/2} (1+\sin^2 x) \, dx$. By using the trigonometric identity $\sin^2 x = \frac{1-\cos 2x}{2}$, the integral becomes $\int_{0}^{\pi/2} (1 + \frac{1-\cos 2x}{2}) \, dx$, which simplifies to $\int_{0}^{\pi/2} (\frac{3}{2} - \frac{\cos 2x}{2}) \, dx$. Performing the integration yields $\left[ \frac{3x}{2} - \frac{\sin 2x}{4} \right]_0^{\pi/2}$, and substituting the limits gives $\frac{3}{2}(\frac{\pi}{2}) - 0$, which results in $\frac{3\pi}{4}$.

Since the initial expression yields the indeterminate form $\frac{0}{0}$, we apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $x$. Using the Fundamental Theorem of Calculus combined with the chain rule, the derivative of the integral $\int_{0}^{x^{2}} (\sin \sqrt{t}) \, dt$ is $\sin(\sqrt{x^{2}}) \cdot 2x$, which simplifies to $2x \sin x$, while the derivative of $x^{3}$ is $3x^{2}$. Dividing these results gives the limit $\lim_{x \rightarrow 0} \frac{2x \sin x}{3x^{2}}$, which simplifies further to $\frac{2}{3} \lim_{x \rightarrow 0} \frac{\sin x}{x}$. Because $\lim_{x \rightarrow 0} \frac{\sin x}{x}$ is a standard limit equal to $1$, the final value is $\frac{2}{3}$.

The property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$ serves as a powerful tool for evaluating definite integrals when the sum of the boundaries allows for algebraic simplification. In this specific case, the limits $\frac{\pi}{6}$ and $\frac{\pi}{3}$ sum to $\frac{\pi}{2}$, which invites the use of the trigonometric identities $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$. By replacing $x$ with $\frac{\pi}{2}-x$ throughout the integrand, the function $g(\alpha)$ transforms into an integral where the numerator becomes $\cos^{\alpha} x$ while the denominator remains unchanged. Adding this new expression to the original integral yields $2g(\alpha) = \int_{\pi/6}^{\pi/3} 1 \, dx$, which evaluates simply to $\frac{\pi}{6}$. Consequently, $g(\alpha)$ simplifies to the constant value $\frac{\pi}{12}$ for any real $\alpha$. Because this result is independent of the input variable, the function maintains the same value for both positive and negative inputs, thereby satisfying the condition $g(\alpha) = g(-\alpha)$ that defines an even function.

Since $\sin x$ is independent of the integration variable $y$, we can treat it as a constant coefficient and rewrite the expression as $f(x) = x + \sin x \int_{0}^{\frac{\pi}{2}} \cos y f(y) dy$. Letting the integral represent a constant $K$, the equation simplifies to $f(x) = x + K \sin x$, which we substitute into the definition of $K$ to obtain:
$K = \int_{0}^{\frac{\pi}{2}} \cos y (y + K \sin y) dy$
Evaluating this integral involves calculating two parts, $\int_{0}^{\frac{\pi}{2}} y \cos y dy$ and $\int_{0}^{\frac{\pi}{2}} K \sin y \cos y dy$, which results in the algebraic relation $K = \frac{\pi-2}{2} + \frac{K}{2}$. Solving this equality reveals that $K = \pi - 2$, which defines the function as $f(x) = x + (\pi - 2) \sin x$.

The function $\log_{e}(\sqrt{1-x}+\sqrt{1+x})$ is an even function, which allows us to simplify the integral over the symmetric interval $[-1, 1]$ to $2 \int_{0}^{1} \log_{e}(\sqrt{1-x}+\sqrt{1+x}) dx$. Applying integration by parts by setting the logarithm as the first function and $1$ as the second, the boundary evaluation at $1$ and $0$ yields $2 \ln(\sqrt{2}) - 0$, which simplifies to $\log_{e} 2$. The derivative of the logarithmic component, combined with rationalization of the resulting expression, transforms the remaining integral into $\int_{0}^{1} \frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}} dx$. Splitting this integral, we calculate $\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} dx - \int_{0}^{1} 1 dx$, which evaluates to $\arcsin(1) - \arcsin(0) - 1$, resulting in $\frac{\pi}{2} - 1$. Summing these parts, the final value of the integral is $\log_{e} 2 + \frac{\pi}{2} - 1$.