Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The greatest integer function is piecewise constant, changing its value only when the expression inside the brackets reaches an integer. To determine these critical transition points, we set the expression $\sqrt{ \frac{10x}{x+1} }$ equal to successive integers $k$, which implies solving the equation $\frac{10x}{x+1} = k^2$ for $x$.

Setting the function equal to $1$, $4$, and $9$ provides the specific boundaries for the integral. Solving $\frac{10x}{x+1} = 1$ yields $x = \frac{1}{9}$, setting it equal to $4$ yields $x = \frac{2}{3}$, and equating it to $9$ leads to $x = 9$. Because the value inside the square root remains between integers within these intervals, the floor function takes on the constant values of $0$, $1$, and $2$ respectively over the segments $[0, \frac{1}{9})$, $[\frac{1}{9}, \frac{2}{3})$, and $[\frac{2}{3}, 9]$.

Calculating the total integral requires splitting the expression into these specific intervals and evaluating the sum $9 \left( \int_{0}^{\frac{1}{9}} 0 \,dx + \int_{\frac{1}{9}}^{\frac{2}{3}} 1 \,dx + \int_{\frac{2}{3}}^{9} 2 \,dx \right)$. Computing the area for each part results in $9 \left( 0 + (\frac{2}{3} - \frac{1}{9}) + 2(9 - \frac{2}{3}) \right)$. Simplifying this expression inside the parentheses gives $\frac{5}{9} + \frac{50}{3}$, which evaluates to $\frac{155}{9}$, and multiplying by $9$ produces the final value of $155$.

To evaluate this limit, we identify it as an indeterminate form of type $\frac{0}{0}$, since the limits of integration converge to the same value as $x$ approaches $\frac{\pi}{2}$. Applying L'Hôpital's rule requires differentiating the numerator and the denominator with respect to $x$. According to the Leibniz integral rule, the derivative of the given integral with respect to $x$ yields $-3x^2 \cos x$, while the derivative of the denominator $(x-\frac{\pi}{2})^2$ is $2(x-\frac{\pi}{2})$.

The limit expression simplifies to $\lim_{x \to \frac{\pi}{2}} \frac{-3x^2 \cos x}{2(x-\frac{\pi}{2})}$. By introducing the substitution $h = x - \frac{\pi}{2}$, we observe that as $x \to \frac{\pi}{2}$, $h \to 0$. Given that $\cos x$ transforms to $\cos(h + \frac{\pi}{2}) = -\sin h$, the expression becomes $\lim_{h \to 0} \frac{3(h + \frac{\pi}{2})^2 \sin h}{2h}$. Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$, the expression reduces to $\frac{3(\frac{\pi}{2})^2}{2}$, which results in $\frac{3\pi^2}{8}$.

Determining the value of this definite integral is most efficiently handled using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$. If we define the integrand as $f(x) = (2x^3 - 3x^2 - x + 1)^{1/3}$, substituting $1-x$ for $x$ transforms the expression into $(2(1-x)^3 - 3(1-x)^2 - (1-x) + 1)^{1/3}$, which simplifies algebraically to $-(2x^3 - 3x^2 - x + 1)^{1/3}$, or $-f(x)$. Because the integral $I$ must be equal to $\int_{0}^{1} f(1-x) \, dx$, and we have determined $f(1-x) = -f(x)$, we find that $I = -I$, which implies the value of the integral is 0.

Integrating $I_k = \int_0^1 (1-x^7)^k dx$ by parts, treating the integrand as the product of $1$ and $(1-x^7)^k$, yields $I_k = 7k \int_0^1 x^7(1-x^7)^{k-1} dx$. Recognizing that $x^7$ can be expressed as $-(1-x^7) + 1$, the integral transforms into $I_k = 7k(I_{k-1} - I_k)$, which leads to the recurrence relation $I_k(1+7k) = 7k I_{k-1}$. Shifting the index by replacing $k$ with $k+1$ in this relation, the ratio $r_k = \frac{I_k}{I_{k+1}}$ is given by:

$r_k = \frac{7k+8}{7k+7}$

Subtracting $1$ from this expression results in $r_k - 1 = \frac{1}{7k+7}$, which simplifies to $7(r_k - 1) = \frac{1}{k+1}$. Consequently, the term inside the summation becomes the reciprocal of this value:

$\sum_{k=1}^{10} \frac{1}{7(r_k - 1)} = \sum_{k=1}^{10} (k+1)$

Calculating the sum of this arithmetic progression from $k=1$ to $10$ provides $2 + 3 + \dots + 11$, which evaluates to $65$.

The definite integral property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ serves as the essential technique for simplifying integrals with limits from $0$ to $\pi$. Using this approach, the integral $I = \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x+\cos^4 x} dx$ is evaluated over the interval $[0, \pi/2]$ by combining $f(x)$ with its transformed counterpart $f(\pi-x)$, which utilizes the anti-symmetry of $\sin(\pi-x) \cos(\pi-x) = -\sin x \cos x$. This summation yields the expression $\int_0^{\pi/2} \frac{(x^2 - (\pi-x)^2) \sin x \cos x}{\sin^4 x+\cos^4 x} dx$, where the numerator simplifies to $2\pi x - \pi^2$. Splitting this into two parts gives $2\pi \int_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x+\cos^4 x} dx - \pi^2 \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x+\cos^4 x} dx$. Applying the same integral property again to the first component cancels the $x$ variable, leaving a single trigonometric integral to be solved using the substitution $\cos 2x = t$. The final evaluation of this expression is $-\frac{\pi^3}{8}$, which, when multiplied by the constant factor $\frac{120}{\pi^3}$ and processed within the absolute value, results in 15.

The most effective approach for evaluating definite integrals with symmetric limits, especially those containing exponential terms in the denominator, utilizes the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. By applying this transformation where $x$ is replaced with $-x$, the original expression $I = \int_{-1}^{1} \frac{\cos \alpha x}{1+3^x} \, dx$ transforms into $I = \int_{-1}^{1} \frac{3^x \cos \alpha x}{3^x+1} \, dx$. Combining these two forms leads to a sum where the numerators share a common denominator, simplifying the integrand significantly to $2I = \int_{-1}^{1} \cos \alpha x \, dx$.

Because $\cos \alpha x$ is an even function, the integral over the interval from $-1$ to $1$ is equivalent to $2 \int_{0}^{1} \cos \alpha x \, dx$. Calculating this integral gives $2 \left[ \frac{\sin \alpha x}{\alpha} \right]_0^1$, which simplifies to $2 \frac{\sin \alpha}{\alpha}$. Dividing both sides by $2$ establishes that $I = \frac{\sin \alpha}{\alpha}$. Equating this result to the provided value of $\frac{2}{\pi}$ confirms that $\alpha = \frac{\pi}{2}$.

The property of definite integrals of odd functions over symmetric limits provides the foundation for this problem, as any integral of an odd function from $-a$ to $a$ is zero. First, verify the parity of $f(x)$ by replacing $x$ with $-x$ and performing a variable substitution $t = -u$, which demonstrates that $f(-x) = -f(x)$, confirming $f(x)$ is indeed an odd function. Since the integration interval spans from $-\pi/2$ to $\pi/2$, the contribution of $f(x)$ vanishes, simplifying the expression to the integral of $\frac{x^2 \cos x}{1+e^x}$.

By applying the property $\int_{-a}^{a} F(x) \, dx = \int_{0}^{a} (F(x) + F(-x)) \, dx$, the integrand transforms remarkably well. Letting $F(x) = \frac{x^2 \cos x}{1+e^x}$, the sum $F(x) + F(-x)$ becomes $\frac{x^2 \cos x}{1+e^x} + \frac{x^2 e^x \cos x}{1+e^x}$, which simplifies to $x^2 \cos x$. Integrating this reduced form from $0$ to $\pi/2$ requires integration by parts. The evaluation yields $[x^2 \sin x]_0^{\pi/2} - 2\int_{0}^{\pi/2} x \sin x \, dx$. Solving the remaining integral results in $\frac{\pi^2}{4} - 2(1)$, which simplifies to $\frac{\pi^2}{4} - 2$. Comparing this result to the given expression $(\frac{\pi}{\alpha})^2 - \alpha$ confirms that $\alpha = 2$.

The integral $\int_{0}^{\pi} \frac{dx}{1-2 a \cos x + a^2}$ hinges on the algebraic symmetry of the denominator, which can be decomposed by applying half-angle identities to $\cos x$. By expressing the denominator as $(1-a)^2 \cos^2(\frac{x}{2}) + (1+a)^2 \sin^2(\frac{x}{2})$, one can divide the numerator and denominator by $\cos^2(\frac{x}{2})$ to introduce the differential form associated with the tangent function. Substituting $u = \tan(\frac{x}{2})$, where $du = \frac{1}{2} \sec^2(\frac{x}{2}) dx$, transforms the integration variable and limits effectively.

This substitution recasts the problem into a standard integral form:
$2 \int_{0}^{\infty } \frac{du}{(1-a)^2 + (1+a)^2 u^2} = \frac{2}{(1+a)^2} \int_{0}^{\infty } \frac{du}{(\frac{1-a}{1+a})^2 + u^2}$
Evaluating this using the result $\int \frac{du}{c^2 + u^2} = \frac{1}{c} \arctan(\frac{u}{c})$, the integration yields $\frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} \cdot \frac{\pi}{2}$. Simplifying the constants leads directly to the final value $\frac{\pi}{1-a^2}$.

The function $f(x) = \frac{4^x}{4^x+2}$ satisfies the symmetry property $f(x) + f(1-x) = 1$. Because the integration limits $f(a)$ and $f(1-a)$ sum to $1$, we can utilize the definite integral property $\int_A^B g(x) \, dx = \int_A^B g(A+B-x) \, dx$. Applying this to the integral
$M = \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) \, dx$
allows us to replace $x$ with $1-x$, yielding $M = \int_{f(a)}^{f(1-a)} (1-x) \sin^4((1-x)x) \, dx$.

Expanding this transformed integral results in $M = \int_{f(a)}^{f(1-a)} \sin^4(x(1-x)) \, dx - \int_{f(a)}^{f(1-a)} x \sin^4(x(1-x)) \, dx$, which simplifies to $M = N - M$. Rearranging this expression gives the relationship $2M = N$. Comparing this result to the given equation $\alpha M = \beta N$, we identify $\alpha = 2$ and $\beta = 1$ as the smallest natural numbers satisfying the condition. Calculating $\alpha^2 + \beta^2$ yields $2^2 + 1^2 = 5$.

The problem is an application of the definite integral as a limit of a Riemann sum, where we convert a summation into an integral of the form $\int_0^1 f(x) dx$. By rewriting the general term $\frac{n^3}{(n^{2+}k^2)(n^{2+}3k^2)}$ as $\frac{1}{n} \cdot \frac{1}{(1+(k/n)^2)(1+3(k/n)^2)}$, the limit as $n \to \infty$ of the summation $\sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{(1+(k/n)^2)(1+3(k/n)^2)}$ simplifies to the integral $\int_0^1 \frac{1}{(1+x^2)(1+3x^2)} dx$.

We decompose the integrand using partial fractions to isolate terms that are easily integrable, which yields the expression $\frac{3/2}{1+3x^2} - \frac{1/2}{1+x^2}$. Integrating each term over the interval from 0 to 1 involves using the standard arctangent integral formula, where $\int \frac{1}{1+a^2x^2} dx = \frac{1}{a} \tan^{-1}(ax) + C$. Applying this to the first term results in $\frac{3}{2} \cdot \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}x)$, and the second term results in $\frac{1}{2} \tan^{-1}(x)$.

Evaluating these components from 0 to 1 gives $\frac{\sqrt{3}}{2} \tan^{-1}(\sqrt{3}) - \frac{1}{2} \tan^{-1}(1)$, which simplifies to $\frac{\sqrt{3}}{2} \cdot \frac{\pi}{3} - \frac{1}{2} \cdot \frac{\pi}{4}$. This evaluates to the value $\frac{\pi}{2\sqrt{3}} - \frac{\pi}{8}$. Through algebraic simplification and rationalization, this result is equivalent to $\frac{13\pi}{8(4\sqrt{3}+3)}$.

The evaluation of this definite integral relies on the symmetry property of integrals, which allows us to simplify the expression by replacing the variable with the sum of the limits minus the variable. We begin by substituting $t = 2x$, which adjusts the differential $dx$ to $\frac{1}{2} dt$ and shifts the upper integration limit to $\frac{\pi}{2}$, resulting in $I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t \, dt}{\sin^4 t + \cos^4 t}$. Applying the property $\int_{0}^{a} f(t) \, dt = \int_{0}^{a} f(a-t) \, dt$ replaces the variable $t$ in the numerator with $\frac{\pi}{2} - t$, allowing the two resulting integrals to be combined to eliminate the variable $t$ from the numerator entirely.

This combination yields $2I = \frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}$. To resolve the remaining integral, we divide the numerator and denominator by $\cos^4 t$, introducing $\sec^4 t$, which is equivalent to $(1 + \tan^2 t) \sec^2 t$. Making the substitution $u = \tan t$ transforms the interval to $[0, \infty )$ and gives $\int_{0}^{\infty } \frac{1+u^2}{u^{4+}1} \, du$. Further simplification by dividing the integrand by $u^2$ allows us to substitute $p = u - \frac{1}{u}$, which results in $\int_{-\infty }^{\infty } \frac{dp}{p^2 + (\sqrt{2})^2}$. Evaluating this integral yields \frac{1}{\sqrt{2}} \[\tan^{-1}(\frac{p}{\sqrt{2}})]_{-∞}^{∞}$, which equals$\frac{\pi}{\sqrt{2}}$. Multiplying this result by the constant$\frac{\pi}{8}$gives$2I = \frac{\pi^2}{8\sqrt{2}}$, which simplifies to$I = \frac{\sqrt{2} \pi^2}{32}$.

The orthocentre $(a, b)$ of the triangle with vertices $(1,2), (2,3)$, and $(3,1)$ lies on the line defined by $x+y=4$, which establishes that the sum of the limits $a+b$ is exactly 4. This geometric property is crucial because it allows us to utilize the definite integral property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ to evaluate the given integrals. By applying this to $I_1 = \int_{a}^{b} x \sin(4x-x^2) dx$, we replace $x$ with $(4-x)$ and note that the argument of the sine function remains unchanged because $4(4-x)-(4-x)^2 = 16-4x-(16-8x+x^2) = 4x-x^2$.

Consequently, $I_1$ can be expressed as $\int_{a}^{b} (4-x) \sin(4x-x^2) dx$, which simplifies to $4 \int_{a}^{b} \sin(4x-x^2) dx - \int_{a}^{b} x \sin(4x-x^2) dx$. Since the first part of this expression is $4I_2$ and the second part is $I_1$, we arrive at the equation $I_1 = 4I_2 - I_1$. Rearranging this gives $2I_1 = 4I_2$, which simplifies to the ratio $\frac{I_1}{I_2} = 2$. Multiplying this ratio by 36 results in 72.

Evaluating this definite integral depends on a strategic substitution that removes the square root term. By defining a new variable $t$ such that $e^x - 1 = t^2$, the integrand converts into a standard form that relates directly to the inverse tangent function, making the integration process straightforward.

Applying the substitution $e^x - 1 = t^2$ yields $e^x dx = 2t dt$, which simplifies the expression $\frac{dx}{\sqrt{e^x - 1}}$ to $\frac{2 dt}{t^2 + 1}$. The integration limits transition from $\alpha$ and $\log_e 4$ to the new values $\sqrt{e^{\alpha} - 1}$ and $\sqrt{e^{\log_e 4} - 1} = \sqrt{3}$. Evaluating the definite integral 2\[\arctan(\sqrt{3}) - \arctan(\sqrt{e^{\alpha} - 1})]$ = \pi / 6$leads to$\pi / 3 - \arctan(\sqrt{e^{\alpha} - 1}) = \pi / 12$, which simplifies to$\arctan(\sqrt{e^{\alpha} - 1}) = \pi / 4$. Consequently,$\sqrt{e^{\alpha} - 1} = 1$, identifying$e^{\alpha} = 2$and$e^{-\alpha} = 1/2$. Constructing a quadratic equation with these roots,$x^2 - (2 + 1/2)x + (2 \cdot 1/2) = 0$, results in$x^2 - 5/2x + 1 = 0$, which rearranges to

$2x^2 - 5x + 2 = 0$

Evaluating integrals of floor functions requires decomposing the integration interval based on the points where the integrand changes its value. For the integral $\int_{0}^{3} \lfloor x^{2} \rfloor dx$, the integrand remains constant between successive square roots of integers, specifically partitioning the interval $(0, 3)$ at $\sqrt{1}, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}$. Summing the areas of these rectangular regions, where each height corresponds to the integer value of $x^2$ in that interval, yields $21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}$.

Similarly, for the integral $\int_{0}^{3} \lfloor \frac{x^{2}}{2} \rfloor dx$, we partition the interval based on where $\frac{x^{2}}{2}$ reaches integer values, occurring at $x = \sqrt{2}, 2, \sqrt{6}, \sqrt{8}$, and the endpoint $3$. Computing the area of these segments gives $10 - 3\sqrt{2} - \sqrt{6}$. Adding these two results together, we obtain the expression $31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7}$. By comparing this to the given form $a + b \sqrt{2} - \sqrt{3} - \sqrt{5} + c \sqrt{6} - \sqrt{7}$, we identify $a = 31$, $b = -6$, and $c = -2$. The sum $a + b + c$ is therefore $31 - 6 - 2$, which equals 23.

The integral of the function $h(x)$ over the range from $-2$ to $2$ is computed by partitioning the region into two parts, represented as $\int_{-2}^{2} h(x) \, dx = \int_{-2}^{0} h(x) \, dx + \int_{0}^{2} h(x) \, dx$. Evaluating these components shows that $\int_{-2}^{0} h(x) \, dx = 2$ and $\int_{0}^{2} h(x) \, dx = 0$. Combining these sums leads to the final result of 2.

Recognizing the pattern of function iteration simplifies the composition process significantly. Since $f(f(x)) = \frac{x}{(1+2x^4)^{1/4}}$, applying the function four times recursively results in $g(x) = \frac{x}{(1+4x^4)^{1/4}}$.

To evaluate the integral $18 \int_{0}^{\sqrt{2\sqrt{5}}} x^2 g(x) dx$, substitute the expression for $g(x)$ to obtain $18 \int_{0}^{\sqrt{2\sqrt{5}}} \frac{x^3}{(1+4x^4)^{1/4}} dx$. Using the substitution $t^4 = 1+4x^4$ leads to the differential relation $4t^3 dt = 16x^3 dx$, which simplifies to $x^3 dx = \frac{1}{4}t^3 dt$. As $x$ varies from $0$ to $\sqrt{2\sqrt{5}}$, the new variable $t$ varies from $1$ to $3$. Substituting these into the integral yields $\frac{18}{4} \int_{1}^{3} \frac{t^3}{t} dt = \frac{9}{2} \int_{1}^{3} t^2 dt$. Computing the final result, \frac{9}{2} \[\frac{t^3}{3}]_{1}^{3} = \frac{3}{2}(27 - 1) = 39$.

Integration by parts provides the most direct path to solving this problem by treating the integrand as the product of $1$ and the logarithmic function. Setting $u = \log_e(x+\sqrt{x^{2+}1})$ and $dv = dx$ allows us to determine $du = \frac{1}{\sqrt{x^{2+}1}} dx$ and $v = x$, which leads to the evaluation of $[x \log_e(x+\sqrt{x^{2+}1})]_{-1}^{2} - \int_{-1}^{2} \frac{x}{\sqrt{x^{2+}1}} dx$.

Evaluating the boundary terms at the given limits yields $2 \log_e(2+\sqrt{5}) - (-1) \log_e(\sqrt{2}-1)$, which simplifies to $\log_e(9+4\sqrt{5}) + \log_e(\sqrt{2}-1)$. The integral part evaluates to $[\sqrt{x^{2+}1}]_{-1}^{2}$, resulting in $\sqrt{5}-\sqrt{2}$. Subtracting this integral value from the boundary result gives $\log_e(9+4\sqrt{5}) + \log_e(\sqrt{2}-1) - (\sqrt{5}-\sqrt{2})$. Because $\log_e(\sqrt{2}-1) = -\log_e(\sqrt{2}+1)$, we combine these terms to reach the final form $\sqrt{2}-\sqrt{5}+\log_e \left( \frac{9+4\sqrt{5}}{1+\sqrt{2}} \right)$.

The expression $\lim_{x\to 0} \frac{x \int_{0}^{x} f(t) \, dt}{e^{x^2}-1}$ simplifies by noting that the denominator behaves like $x^2$ near zero, specifically $\lim_{x\to 0} \frac{e^{x^2}-1}{x^2} = 1$. By dividing both the numerator and denominator by $x^2$, the entire expression can be rewritten as the quotient of two limits: $\lim_{x\to 0} \frac{\int_{0}^{x} f(t) \, dt}{x}$ divided by $\lim_{x\to 0} \frac{e^{x^2}-1}{x^2}$.

The denominator limit is $1$. The numerator represents an indeterminate form of type $\frac{0}{0}$, which is resolved using L'Hospital's rule by differentiating the integral in the numerator and the $x$ in the denominator, yielding $\lim_{x\to 0} \frac{f(x)}{1} = f(0)$. Since $f(0) = \frac{1}{2}$, the result for $\alpha$ is $\frac{1}{2}$. Consequently, the final evaluation is:
$8\alpha^2 = 8 \left( \frac{1}{2} \right)^2 = 8 \cdot \frac{1}{4} = 2$

The expression $\lim_{x \to 0} \frac{f(x)}{x^3}$ results in an indeterminate $\frac{0}{0}$ form, which allows for the application of L'Hôpital's Rule. By differentiating the numerator using the Fundamental Theorem of Calculus to obtain $f'(x) = x + \sin(1-e^x)$ and differentiating the denominator to obtain $3x^2$, the limit simplifies to $\lim_{x \to 0} \frac{x + \sin(1-e^x)}{3x^2}$.

Because this ratio remains in the $\frac{0}{0}$ form, differentiating a second time yields $\lim_{x \to 0} \frac{1 - e^x \cos(1-e^x)}{6x}$. Applying L'Hôpital's Rule one final time requires differentiating the numerator to obtain $-(e^x \cos(1-e^x) + e^{2x} \sin(1-e^x))$ and the denominator to obtain $6$. Evaluating this expression as $x$ approaches $0$ results in $\frac{-(1 \cdot 1 + 1 \cdot 0)}{6}$, which equals $-\frac{1}{6}$.