WB JEE 2011 | Application of Integration Question 7 | Mathematics

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WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The area enclosed between y² = x and y = x is

$${2 \over 3}$$ sq. units

$${1 \over 2}$$ sq. units

$${1 \over 3}$$ sq. units

$${1 \over 6}$$ sq. units

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The area bounded by y² = 4x and x² = 4y is

$${{20} \over 3}$$ sq. units

$${{16} \over 3}$$ sq. units

$${{14} \over 3}$$ sq. units

$${{10} \over 3}$$ sq. units

WB JEE 2026

MCQ (Single Correct Answer)

-0.25

Consider the function $y=f(x)$ defined implicitly by the equation $y^3-3 y+x=0$ on the interval $(-\infty,-2) \cup(2, \infty)$. The area of the region bounded by the curve $y=f(x)$, the $x$-axis and the lines $x=a, x=b$, where $-\infty< a< b< -2$ is

$\int_a^b \frac{x d x}{3\left((f(x))^2-1\right)}-b f(b)+a f(a)$

$\int_a^b \frac{x d x}{3\left((f(x))^2-1\right)}+b f(b)-a f(a)$

$\quad-\int_a^b \frac{x d x}{3\left((f(x))^2-1\right)}-b f(b)+a f(a)$

$\quad-\int_a^b \frac{x d x}{3\left((f(x))^2-1\right)}+b f(b)-a f(a)$

WB JEE 2024

MCQ (Single Correct Answer)

-0.25

The area bounded by the curves $$x=4-y^2$$ and the Y-axis is

16 sq. unit

$$\frac{32}{3}$$ sq. unit

$$\frac{16}{3}$$ sq. unit

32 sq. unit

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