The value of =120(|sqrt(pi(0)^rx| x|dx)|) is ____. — jee_main_2026_28_january

More Integrals Previous-Year Questions — Page 4

Q13 jee_main_2026_28_january_evening Greatest Integer Function integrals

Let

[\cdot]

denote the greatest integer function. Then

\int-(π)/(2)(π)/(2)((12(3+[x]))/(3+[ x]+[ x]))dx

is equal to:

A. $15π + 4$ $15\pi + 4$
B. $11π + 2$ $11\pi + 2$
C. $13π + 1$ $13\pi + 1$
D. $12π + 5$ $12\pi + 5$

Solution

Core Logic

Split the integral $I$ $I$ at integer points and trigonometric step thresholds between $-π/2 \approx -1.57$ $-\pi/2 \approx -1.57$ and $π/2 \approx 1.57$ $\pi/2 \approx 1.57$: Intervals are: $(-π/2, -1), (-1, 0), (0, 1), (1, π/2)$ . For all $x in (-π/2, π/2)$ $x \in (-\pi/2, \pi/2)$ excluding exactly $0$ $0$, $[ x] = 0$ $[\cos x] = 0$. $[ x] = -1$ $[\sin x] = -1$ for $x in (-π/2, 0)$ $x \in (-\pi/2, 0)$ and $[ x] = 0$ $[\sin x] = 0$ for $x in (0, π/2)$ $x \in (0, \pi/2)$.

Execution

Evaluate in pieces:

$x in (-π/2, -1)$ $x \in (-\pi/2, -1)$:

$[x] = -2, [ x] = -1, [ x] = 0$ $[x] = -2, [\sin x] = -1, [\cos x] = 0$. Integrand = $(12(3 - 2))/(3 - 1 + 0) = (12(1))/(2) = 6$ .

$x in (-1, 0)$ $x \in (-1, 0)$:

$[x] = -1, [ x] = -1, [ x] = 0$ $[x] = -1, [\sin x] = -1, [\cos x] = 0$. Integrand = $(12(3 - 1))/(2) = (24)/(2) = 12$ .

$x in (0, 1)$ $x \in (0, 1)$:

$[x] = 0, [ x] = 0, [ x] = 0$ $[x] = 0, [\sin x] = 0, [\cos x] = 0$. Integrand = $(12(3 + 0))/(3) = 12$ $\frac{12(3 + 0)}{3} = 12$.

$x in (1, π/2)$ $x \in (1, \pi/2)$:

$[x] = 1, [ x] = 0, [ x] = 0$ $[x] = 1, [\sin x] = 0, [\cos x] = 0$. Integrand = $(12(3 + 1))/(3) = (48)/(3) = 16$ .

Integrate piece by piece:

I = \int-π/2⁻¹ 6 dx + \int₋₁⁰ 12 dx + \int₀¹ 12 dx + \int₁π/2 16 dx

I = 6(-1 + π/2) + 12(0 - (-1)) + 12(1 - 0) + 16(π/2 - 1)

I = 3π - 6 + 12 + 12 + 8π - 16

$I = 11π + 2$ $I = 11\pi + 2$

Pattern Recognition

For integrals combining integer functions and trigonometric boundaries, breaking the domain purely at integer values $(-1, 0, 1)$ $(-1, 0, 1)$ generally perfectly matches trigonometric boundaries since the ranges of $\sin$ and $\cos$ in $(-π/2, π/2)$ $(-\pi/2, \pi/2)$ hover near these integers.

Chapter Mix

Class 12 Maths: Definite Integration

Q19 jee_main_2026_28_january_evening Substitution of Fractional Powers

Let

f(x) = \int dxx(2/3) + 2x(1/2)

be such that

f(0) = -26 + 24 ₑ(2)

. If

f(1) = a + b ₑ(3)

where

a, b in Z

, then

a + b

is equal to:

A. $-18$ $-18$
B. $-5$ $-5$
C. $-11$ $-11$
D. $-26$ $-26$

Solution

Core Logic

f(x) = \int dxx2/3 + 2x1/2

To eliminate fractional powers, use substitution. The LCM of denominators 3 and 2 is 6. Put $x = t⁶ \Rightarrow dx = 6t⁵ dt$ $x = t^6 \Rightarrow dx = 6t^5 \, dt$.

Execution

f(x) = \int (6t⁵ dt)/(t⁴ + 2t³) = 6 \int (t²)/(t + 2) dt

Rewrite the numerator using polynomial division or algebraic manipulation:

6 \int ((t² - 4) + 4)/(t + 2) dt = 6 [ \int (t - 2) dt + 4 \int (1)/(t + 2) dt ]

= 6 [ (t²)/(2) - 2t + 4 ln|t + 2| ] + C

= 3t² - 12t + 24 ln|t + 2| + C

Substitute $t = x1/6$ $t = x^{1/6}$:

f(x) = 3x1/3 - 12x1/6 + 24 ln|x1/6 + 2| + C

Evaluate constant $C$ $C$ using $f(0)$ $f(0)$: $f(0) = 24 ln 2 + C = -26 + 24 ln 2 \Rightarrow C = -26$ .

Find $f(1)$ $f(1)$:

f(1) = 3(1) - 12(1) + 24 ln(1 + 2) - 26 = -9 - 26 + 24 ln 3 = -35 + 24 ln 3

Given $f(1) = a + b ln 3$ $f(1) = a + b \ln 3$, we map $a = -35$ $a = -35$ and $b = 24$ $b = 24$. $a + b = -35 + 24 = -11$ $a + b = -35 + 24 = -11$.

Pattern Recognition

For integrals with multiple roots $x1/m$ $x^{1/m}$ and $x1/n$ $x^{1/n}$, substituting $x = tLCM(m,n)$ $x = t^{LCM(m,n)}$ strictly normalizes the algebraic fraction into simple polynomial synthetic division.

Chapter Mix

Class 12 Maths: Indefinite Integration

Q57 jee_main_2025_02_april_evening Properties of Definite Integrals

Let

(a, b)

be the point of intersection of the curve

x² = 2y

and the straight line

y - 2x - 6 = 0

in the second quadrant. Then the integral

I = \intₐb (9x²)/(1 + 5^x) dx

is equal to:

A. 24
B. 27
C. 18
D. 21

Solution

Related Formula

King's Property: \intₐb f(x) dx = \intₐb f(a+b-x) dx

Core Logic

First, we find the coordinates of intersection in the second quadrant to determine the integration limits $a$ $a$ and $b$ $b$.

Step 1: Find points of intersection

Substitute $y = (x²)/(2)$ $y = \frac{x^2}{2}$ into the line equation $y - 2x - 6 = 0$ $y - 2x - 6 = 0$:

(x²)/(2) - 2x - 6 = 0 x² - 4x - 12 = 0

(x - 6)(x + 2) = 0 x = 6 or x = -2

Since the point $(a, b)$ $(a, b)$ lies in the second quadrant, $x$ $x$ must be negative:

$a = -2$ $a = -2$

b = 2(a) + 6 = 2(-2) + 6 = 2

Thus, the integration limits are $a = -2$ $a = -2$ and $b = 2$ $b = 2$.

Step 2: Solve the Integral using King's Property

The integral is:

I = \int₋₂² (9x²)/(1 + 5^x) dx --- (1)

Apply King's property, substituting $x \to -x$ $x \to -x$ (since $-2 + 2 - x = -x$ $-2 + 2 - x = -x$):

I = \int₋₂² 9(-x)²1 + 5-x dx = \int₋₂² (9x² \cdot 5^x)/(1 + 5^x) dx --- (2)

Adding equations (1) and (2):

2I = \int₋₂² 9x² ( (1 + 5^x)/(1 + 5^x) ) dx = \int₋₂² 9x² dx

Since $9x²$ $9x^2$ is an even function:

2I = 2 \int₀² 9x² dx I = [ 3x³ ]₀² = 3(8) - 0 = 24

Pattern Recognition

Whenever you see an exponential denominator like $1 + c^x$ $1 + c^x$ inside a symmetric interval definite integral $[-a, a]$ $[-a, a]$, applying King's property will almost always cancel the exponential factor cleanly when the remaining numerator is even.

Chapter Mix

Class 12 Mathematics: Integral Calculus

Q63 jee_main_2025_02_april_evening Integration by Rationalisation

4\int₀¹( 1\sqrt(3 + x²) + \sqrt(1 + x²))dx - 3 ₑ(\sqrt(3))

is equal to:

A. $2 + \sqrt(2) + ₑ(1 + \sqrt(2))$
B. $2 - \sqrt(2) - ₑ(1 + \sqrt(2))$
C. $2 + \sqrt(2) - ₑ(1 + \sqrt(2))$
D. $2 - \sqrt(2) + ₑ(1 + \sqrt(2))$

Solution

Related Formula

\int \sqrt(a² + x²) dx = (x)/(2) \sqrt(a² + x²) + (a²)/(2) ln| x + \sqrt(a² + x²) |

Core Logic

We first rationalise the denominator to split the integral into two standard integration terms.

Step 1: Rationalise the integrand

Multiply the numerator and denominator by $\sqrt(3+x²) - \sqrt(1+x²)$ $\sqrt{3+x^2} - \sqrt{1+x^2}$:

1\sqrt(3 + x²) + \sqrt(1 + x²) = \sqrt(3 + x²) - \sqrt(1 + x²)(3+x²) - (1+x²) = \sqrt(3 + x²) - \sqrt(1 + x²)2

Thus, the integral expression simplifies to:

I = 4 \int₀¹ ( \sqrt(3 + x²) - \sqrt(1 + x²)2 ) dx - 3 ₑ(\sqrt(3))

I = 2 \int₀¹ \sqrt(3 + x²) dx - 2 \int₀¹ \sqrt(1 + x²) dx - (3)/(2) ₑ 3

Step 2: Evaluate the integrals

For the first integral:

2 \int₀¹ \sqrt(3 + x²) dx = 2 [ (x)/(2) \sqrt(3 + x²) + (3)/(2) ln| x + \sqrt(3 + x²) | ]₀¹

= [ x \sqrt(3 + x²) + 3 ln| x + \sqrt(3 + x²) | ]₀¹

= ( \sqrt(4) + 3 ln(1 + \sqrt(4)) ) - ( 0 + 3 ln\sqrt(3) )

= 2 + 3 ln 3 - (3)/(2) ln 3 = 2 + (3)/(2) ln 3

For the second integral:

-2 \int₀¹ \sqrt(1 + x²) dx = -2 [ (x)/(2) \sqrt(1 + x²) + (1)/(2) ln| x + \sqrt(1 + x²) | ]₀¹

= - [ x \sqrt(1 + x²) + ln| x + \sqrt(1 + x²) | ]₀¹

= - ( \sqrt(2) + ln(1 + \sqrt(2)) )

Step 3: Sum the terms

Now compile all terms:

I = ( 2 + (3)/(2) ln 3 ) - \sqrt(2) - ln(1 + \sqrt(2)) - (3)/(2) ln 3

I = 2 - \sqrt(2) - ln(1 + \sqrt(2))

Pattern Recognition

Integration of roots of quadratics: Always look for algebraic rationalisation when dealing with sum of root denominators. It directly reduces complex fractions into standard integrable functions.

Chapter Mix

Class 12 Mathematics: Integral Calculus

The value of Σr=1²⁰(| π(∫₀rx| π x|dx)|)$\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r}x\left|\sin\pi x\right|dx\right)}\right|\right)$ is ____.

Solution & Explanation

Core Logic

Step 1: Integral of Modulus

Step 2: Evaluate the Sum

Chapter Mix

Reference Study Guides

More Integrals Previous-Year Questions — Page 4

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find points of intersection

Step 2: Solve the Integral using King's Property

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Rationalise the integrand

Step 2: Evaluate the integrals

Step 3: Sum the terms

Pattern Recognition

Chapter Mix

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The value of $Σr=1²⁰(| π(\int₀rx| π x|dx)|)$ is ____.