Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

This expression represents a Riemann sum that converges to a definite integral as $n$ approaches infinity. The general term, written as $\frac{1}{n} \cdot \left( \frac{r}{n} \right) \sec^2 \left( \frac{r}{n} \right)^2$, clearly maps to the integral $\int_{0}^{1} x \sec^2 (x^2) \, dx$.

Applying the substitution $u = x^2$ leads to $du = 2x \, dx$, which allows the integral to be rewritten as $\frac{1}{2} \int_{0}^{1} \sec^2 u \, du$. Because the antiderivative of $\sec^2 u$ is $\tan u$, evaluating this expression across the bounds from $0$ to $1$ results in $\frac{1}{2} \tan 1$.

To evaluate this integral for $a>0$, we apply the symmetry property $\int_{-k}^{k} f(x) \, dx = \int_{-k}^{k} f(-x) \, dx$. Letting $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} \, dx$, substituting $-x$ yields $I = \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{1+a^x} \, dx$ after accounting for the identity $a^{-x} = \frac{1}{a^x}$. Summing both expressions for $I$ eliminates the denominator, leaving $2I = \int_{-\pi}^{\pi} \cos^2 x \, dx$, which reduces to $4 \int_{0}^{\pi/2} \cos^2 x \, dx$ due to the properties of even periodic functions. Since the integral $\int_{0}^{\pi/2} \cos^2 x \, dx = \frac{\pi}{4}$, we find $2I = 4 \left( \frac{\pi}{4} \right) = \pi$, which yields $I = \frac{\pi}{2}$.

For definite integrals over the same limits, the integral of a larger integrand is always greater than the integral of a smaller one. On the interval $0 < x < 1/2$, the function $x^2$ remains consistently greater than $x^3$ because $x$ is a positive fraction less than $1$. This establishes that $I_1 = \int_0^{1/2} x^2 dx$ is greater than $I_2 = \int_0^{1/2} x^3 dx$ because the integrand of $I_1$ is larger than that of $I_2$ throughout the integration range.

Definite integrals defined over the range $[a, b]$ often simplify significantly when applying the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$. For the given integral, the sum of the lower limit $3$ and upper limit $6$ equals $9$. Substituting $x$ with $9-x$ in the integrand $\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}$ produces an equivalent integral with the same denominator but a numerator of $\sqrt{9-x}$. Adding these two integral expressions together creates an integrand where the numerator and denominator are identical, reducing the operation to calculating $\int_{3}^{6} 1 \, dx$. The result of this integration is $6 - 3$, which equals $3$, and since the sum of the two identical integrals was $2I$, the value of the integral is half of this, yielding $\frac{3}{2}$.

Because the limit results in an indeterminate $\frac{0}{0}$ form at $x=2$, apply L'Hôpital's rule by differentiating the integral and the denominator with respect to $x$. Differentiating the numerator $\int_6^{f(x)} 4t^3 dt$ using the Leibniz rule yields $4(f(x))^3 f'(x)$, while the derivative of the denominator $x-2$ is 1. Substituting the limit $x \to 2$ gives $4(f(2))^3 f'(2)$, and inserting the given values $f(2)=6$ and $f'(2)=\frac{1}{48}$ leads to the calculation $4 \times 6^3 \times \frac{1}{48}$, which equals 18.

Evaluating a limit of a sum as $n$ approaches infinity involves recognizing it as a definite integral of the form $\int_0^1 f(x) dx$, derived from the Riemann sum definition $\lim_{n\to \infty } \sum_{r=1}^{n} \frac{1}{n} f(\frac{r}{n})$. In this expression, identifying the function $f(\frac{r}{n}) = e^{\frac{r}{n}}$ leads to $f(x) = e^x$, while the limits of integration are determined by the range of $\frac{r}{n}$ as $r$ varies from $1$ to $n$, which spans from $0$ to $1$.

$\lim_{n\to \infty } \sum_{r=1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \int_0^1 e^x dx$

The integral of $e^x$ with respect to $x$ is $e^x$, and applying the fundamental theorem of calculus by evaluating this antiderivative at the upper and lower bounds produces the final result.

$\$[e^x]_0^1 = e^1 - e^0 = e - 1$

Trigonometric identities provide the key to simplifying the expression inside the integral. Since $1 + \sin 2x$ is equivalent to $(\sin x + \cos x)^2$, the denominator simplifies to $\sqrt{(\sin x + \cos x)^2}$, which is just $\sin x + \cos x$ given that both sine and cosine are positive on the interval from $0$ to $\frac{\pi}{2}$.

This simplification reduces the entire fraction to $\sin x + \cos x$. The definite integral is then evaluated as follows:
$\int_{0}^{\pi/2} (\sin x + \cos x) dx = [-\cos x + \sin x]_{0}^{\pi/2}$
Substituting the boundaries of the interval leads to $(-\cos \frac{\pi}{2} + \sin \frac{\pi}{2}) - (-\cos 0 + \sin 0)$. Computing these values, we find $(0 + 1) - (-1 + 0)$, which equals $1 - (-1) = 2$.

Evaluating the integral $\int_{0}^{\pi} x f(\sin x) \, dx$ relies on the property $\int_{0}^{a} g(x) \, dx = \int_{0}^{a} g(a-x) \, dx$. By applying this transformation, the integral becomes $I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx$. Since $\sin(\pi - x) = \sin x$, the expression simplifies to $I = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx$. Adding the original integral $I$ and this transformed version yields $2I = \int_{0}^{\pi} (x + \pi - x) f(\sin x) \, dx$, which simplifies to $\pi \int_{0}^{\pi} f(\sin x) \, dx$. Using the symmetry property $\int_{0}^{2a} g(x) \, dx = 2 \int_{0}^{a} g(x) \, dx$ when the function is symmetric, we can rewrite $\int_{0}^{\pi} f(\sin x) \, dx$ as $2 \int_{0}^{\pi/2} f(\sin x) \, dx$. Substituting this back results in $2I = 2\pi \int_{0}^{\pi/2} f(\sin x) \, dx$, simplifying to $I = \pi \int_{0}^{\pi/2} f(\sin x) \, dx$. Comparing this to the given equation $I = A \int_{0}^{\pi/2} f(\sin x) \, dx$, it is clear that $A = \pi$.

The absolute value function changes sign at the roots of $1-x^2$, requiring the interval $[-2, 3]$ to be split at $x = -1$ and $x = 1$ to evaluate the integral correctly. Because the function $|1-x^2|$ is equivalent to $x^{2-}1$ when $x^{2-}1 \ge 0$ and $1-x^2$ when $1-x^2 \ge 0$, we analyze the behavior on each segment: $1-x^2$ is negative on the outer intervals $[-2, -1]$ and $[1, 3]$, while it is positive on the middle interval $[-1, 1]$.

This leads to the expression $\int_{-2}^{-1} (x^{2-}1) \, dx + \int_{-1}^{1} (1-x^2) \, dx + \int_{1}^{3} (x^{2-}1) \, dx$. Evaluating the first integral gives $[\frac{x^3}{3} - x]_{-2}^{-1} = (-\frac{1}{3} + 1) - (-\frac{8}{3} + 2) = \frac{4}{3}$. For the second interval, the integral is $[x - \frac{x^3}{3}]_{-1}^{1} = (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) = \frac{4}{3}$. Calculating the final part yields $[\frac{x^3}{3} - x]_{1}^{3} = (9 - 3) - (\frac{1}{3} - 1) = 6 + \frac{2}{3} = \frac{20}{3}$. Summing these segments gives the final value of $\frac{28}{3}$.

Definite integrals possess a property stating that $\int_{A}^{B} h(x) \, dx = \int_{A}^{B} h(A+B-x) \, dx$, which allows for simplifying complex expressions through variable substitution. Summing the integral limits $f(a)$ and $f(-a)$ with $f(x) = \frac{e^x}{1+e^x}$ provides $f(a) + f(-a) = \frac{e^a}{1+e^a} + \frac{1}{1+e^a} = 1$. By applying the transformation $x \to 1-x$, the integral $I_1 = \int_{f(-a)}^{f(a)} x g\{x(1-x)\} \, dx$ transforms into $\int_{f(-a)}^{f(a)} (1-x) g\{(1-x)(1-(1-x))\} \, dx$.

Because the argument $(1-x)(1-(1-x))$ simplifies to $x(1-x)$, the integrand $g\{x(1-x)\}$ remains invariant under this transformation. The integral expression becomes $I_1 = \int_{f(-a)}^{f(a)} (1-x) g\{x(1-x)\} \, dx$, which expands into $\int_{f(-a)}^{f(a)} g\{x(1-x)\} \, dx - \int_{f(-a)}^{f(a)} x g\{x(1-x)\} \, dx$. Identifying these components, $\int_{f(-a)}^{f(a)} g\{x(1-x)\} \, dx$ corresponds to $I_2$ and the second term is $I_1$. The equation simplifies to $I_1 = I_2 - I_1$, leading to $2 I_1 = I_2$, which indicates that the ratio $\frac{I_2}{I_1}$ is 2.

Since direct substitution of $x = 0$ yields the indeterminate form $\frac{0}{0}$, applying L'Hôpital's rule allows us to resolve the limit by differentiating the numerator and denominator independently. Using the Leibniz integral rule for the numerator and the product rule for the denominator yields the expression
$\frac{2x \sec^2(x^2)}{\sin x + x \cos x}$
Dividing both the numerator and the denominator by $x$ simplifies the expression to
$\frac{2 \sec^2(x^2)}{\frac{\sin x}{x} + \cos x}$
Evaluating this limit as $x \to 0$ using the standard values $\sec^2(0) = 1$, $\lim_{x \to 0} \frac{\sin x}{x} = 1$, and $\cos 0 = 1$ results in $\frac{2(1)}{1+1}$, which equals 1.

The condition $f'(x)=f(x)$ defines a function where the rate of change is proportional to its current value, characterizing the exponential function $f(x)=e^x$ given the initial condition $f(0)=1$. Since the relationship $f(x)+g(x)=x^2$ holds, we determine $g(x)=x^{2-}e^x$, which leads to the integral expression:
$\int_{0}^{1} e^x(x^{2-}e^x) dx = \int_{0}^{1} x^2 e^x dx - \int_{0}^{1} e^{2x} dx$
Evaluating the first term $\int_{0}^{1} x^2 e^x dx$ using integration by parts yields $e^x(x^{2-}2x+2)$, which results in $e-2$ when evaluated over the interval from $0$ to $1$. The second integral $\int_{0}^{1} e^{2x} dx$ calculates to $\frac{1}{2} e^{2x}$ evaluated from $0$ to $1$, yielding $\frac{e^{2-}1}{2}$. Subtracting these values provides $e-2 - (\frac{e^2}{2} - \frac{1}{2})$, which simplifies to $e - \frac{e^2}{2} - \frac{3}{2}$.

Treating these limits as Riemann sums allows us to represent infinite summations as definite integrals. The first term, $\frac{1+2^{4+}\dots+n^4}{n^5}$, can be rewritten as $\frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^4$, which converges to the definite integral $\int_0^1 x^4 \,dx$. Evaluating this integral gives $\frac{1}{5}$. The second term, $\frac{1+2^{3+}\dots+n^3}{n^5}$, can be broken down into $\frac{1}{n} \cdot \left( \frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^3 \right)$. As $n$ approaches infinity, the summation component converges to $\int_0^1 x^3 \,dx$, which equals $\frac{1}{4}$, while the leading factor of $\frac{1}{n}$ causes the entire second term to approach 0. Subtracting 0 from $\frac{1}{5}$ results in a final value of $\frac{1}{5}$.

Utilizing the definite integral property $\int_{0}^{1} f(x) dx = \int_{0}^{1} f(1-x) dx$ simplifies the integrand significantly. By substituting $x$ with $(1-x)$, the expression $x(1-x)^n$ becomes $(1-x)x^n$, which expands to $x^n - x^{n+1}$. Integration of these terms results in $\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}$. Applying the limits from $0$ to $1$ leaves the final value as $\frac{1}{n+1} - \frac{1}{n+2}$.

This problem focuses on computing the convolution integral $F(t) = \int_0^t f(t-y) g(y) \, dy$. Substituting the specific expressions $f(y) = e^y$ and $g(y) = y$ into this integral allows us to solve for $F(t)$ by integrating with respect to $y$ over the interval from $0$ to $t$.

Substituting the given functions yields $F(t) = \int_0^t e^{t-y} y \, dy$. Factoring out the term $e^t$, which acts as a constant with respect to the variable of integration $y$, simplifies the expression to $F(t) = e^t \int_0^t y e^{-y} \, dy$. Using integration by parts to evaluate $\int y e^{-y} \, dy$, we identify the antiderivative as $-(y+1)e^{-y}$. Applying the fundamental theorem of calculus over the limits $0$ to $t$, we calculate the definite integral as $-(t+1)e^{-t} - (-(0+1)e^0)$, which simplifies to $1 - (t+1)e^{-t}$. Multiplying this result by the leading factor of $e^t$ gives $e^t - e^t(t+1)e^{-t}$, which resolves to $e^t - (1+t)$.

The definite integral property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a+b-x) \, dx$ allows us to rewrite the integral $I = \int_{a}^{b} x f(x) \, dx$ by replacing the variable $x$ with $a+b-x$. Applying this transformation yields:
$I = \int_{a}^{b} (a+b-x) f(a+b-x) \, dx$
Given the symmetry condition $f(a+b-x) = f(x)$, we substitute $f(x)$ into the integrand, which simplifies the expression to $I = \int_{a}^{b} (a+b-x) f(x) \, dx$. Distributing $f(x)$ across the binomial term results in:
$I = (a+b) \int_{a}^{b} f(x) \, dx - \int_{a}^{b} x f(x) \, dx$
Since the second term on the right is identical to the original integral $I$, the expression becomes $I = (a+b) \int_{a}^{b} f(x) \, dx - I$. Adding $I$ to both sides produces $2I = (a+b) \int_{a}^{b} f(x) \, dx$, which confirms the final result:
$I = \frac{a+b}{2} \int_{a}^{b} f(x) \, dx$

The limit $\lim_{n\to \infty }\frac{1^p+2^p+3^p+\dots+n^p}{n^{p+1}}$ represents a Riemann sum that converges to a definite integral as the number of partitions approaches infinity. By distributing the $n^{p+1}$ term in the denominator as $n^p \cdot n$, the expression can be rewritten as:

$\lim_{n\to \infty } \sum_{r=1}^{n} \left(\frac{r}{n}\right)^p \cdot \frac{1}{n} = \int_{0}^{1} x^p \, dx$

Evaluating this integral using the antiderivative $\frac{x^{p+1}}{p+1}$ over the interval from 0 to 1 yields $\frac{1}{p+1}$.

Evaluating an integral over a symmetric range such as $[-\pi, \pi]$ becomes significantly more manageable by classifying the integrand components as odd or even. Expanding the initial expression into two separate parts yields:

$\int_{-\pi}^{\pi} \frac{2x}{1+\cos^{2}x} dx + \int_{-\pi}^{\pi} \frac{2x\sin x}{1+\cos^{2}x} dx$

The first term, $\frac{2x}{1+\cos^{2}x}$, represents an odd function because replacing $x$ with $-x$ negates the entire expression, causing its integral over the symmetric interval $[-\pi, \pi]$ to equal zero. Conversely, the second term, $\frac{2x\sin x}{1+\cos^{2}x}$, is an even function because the product of two odd functions-$x$ and $\sin x$-results in an even function, which allows us to double the integral over the half-interval $[0, \pi]$ to obtain $4 \int_{0}^{\pi} \frac{x\sin x}{1+\cos^{2}x} dx$.

Applying the definite integral property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ to this expression transforms it into $4 \int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\cos^{2}(\pi-x)} dx$. Given that $\sin(\pi-x) = \sin x$ and $\cos^{2}(\pi-x) = \cos^{2}x$, this simplifies to $4 \int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\cos^{2}x} dx$. Distributing the terms results in the equation $I = 4\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^{2}x} dx - I$, where $I$ represents the value of our integral. Rearranging this leads to $2I = 4\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^{2}x} dx$, which reduces to $I = 2\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos^{2}x} dx$.

To calculate this value, we substitute $u = \cos x$, which implies $du = -\sin x dx$, effectively changing the limits of integration from $[0, \pi]$ to $[1, -1]$. The integral becomes:

$I = 2\pi \int_{-1}^{1} \frac{du}{1+u^2} = 2\pi \$[\tan^{-1} u]_{-1}^{1}$

Evaluating this at the boundaries yields $2\pi \left(\frac{\pi}{4} - (-\frac{\pi}{4})\right)$, which simplifies to $2\pi \cdot \frac{\pi}{2}$, resulting in a final value of $\pi^2$.

Integration by parts serves as the essential method for evaluating expressions like $\int_{0}^{2} x f'(x) \, dx$ by splitting the integrand into a product of a variable and a derivative. Following the standard formula $\int u \, dv = uv - \int v \, du$, we set $u = x$ and $dv = f'(x) \, dx$, which leads to $du = dx$ and $v = f(x)$, resulting in the following transformation:
$\int_{0}^{2} x f'(x) \, dx = [x f(x)]_0^2 - \int_{0}^{2} f(x) \, dx$
Evaluating the boundary term $[x f(x)]_0^2$ results in $2 f(2) - 0 f(0)$, and since the function intersects the x-axis at 2, we have $f(2) = 0$, meaning this term simplifies to zero. Given that the area enclosed by the function and the axes is $\frac{3}{4}$, the integral simplifies to $0 - \frac{3}{4}$, which yields $-\frac{3}{4}$.

The absolute value function $|\sin x|$ is periodic with a period of $\pi$, meaning the area under the curve repeats identically for every interval of length $\pi$. Since $\sin x$ is positive within the range from $0$ to $\pi$, we can calculate the area of one such arch by evaluating $\int_{0}^{\pi} \sin x dx$. The antiderivative $[-\cos x]_{0}^{\pi}$ yields $-(\cos \pi - \cos 0)$, which simplifies to $-(-1 - 1)$ and results in 2. Because the total integration interval extends from 0 to $10\pi$, covering exactly 10 of these identical periods, we find the total area by calculating the product of the number of intervals and the area of one period.

$10 \times 2 = 20$