List of top Mathematics Questions on Area under Simple Curves

The area of the region enclosed between the circles $x^2 + y^2 = 4$ and $x^2 + (y - 2)^2 = 4$ is:

The area of the region, inside the circle $(x-2\sqrt{3})^2 + y^2 = 12$ and outside the parabola $y^2 = 2\sqrt{3}x$ is:

The area of the region enclosed between the curve $ y = |x| $, x-axis, $ x = -2 $} and $ x = 2 $ is:

If the area of the region \[ \{(x, y) : |4 - x^2| \leq y \leq x^2, y \leq 4, x \geq 0\} \] is $ \frac{80\sqrt{2}}{\alpha - \beta} $, where $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:

Let the area enclosed between the curves $ |y| = 1 - x^2 $ and $ x^2 + y^2 = 1 $ be $ \alpha $. If $ 9\alpha = \beta\pi + \gamma $; $ \beta, \gamma $ are integers, then the value of $ |\beta - \gamma| $ equals:

Let the area of the region $ \{(x, y) : 2y \leq x^2 + 3, \, y + |x| \leq 3, \, y \geq |x - 1|\} $ be $ A $. Then $ 6A $ is equal to:

What is the area of a triangle with base 8 cm and height 6 cm?

The area of the region in the first quadrant inside the circle $x^2 + y^2 = 8$ and outside the parabola $y^2 = 2x$ is equal to:

If the area of the region \[ \left\{(x, y) : \frac{a}{x^2} \leq y \leq \frac{1}{x}, \, 1 \leq x \leq 2, \, 0<a<1 \right\} \] is \[ (\log 2) - \frac{1}{7}, \] then the value of $7a - 3$ is equal to:

The area enclosed by the curves xy + 4y = 16 and x + y = 6 is equal to :

The area enclosed between the curves $y = x|x|$ and $y = x - |x|$ is:

Let the area of the region enclosed by the curve $y = \min\{\sin x, \cos x\}$ and the x-axis between $x = -\pi$ to $x = \pi$ be $A$. Then $A^2$ is equal to _____.

One of the points of intersection of the curves $ y = 1 + 3x - 2x^2 $ and $ y = \frac{1}{x} $ is $ \left( \frac{1}{2}, 2 \right) $. Let the area of the region enclosed by these curves be \[\frac{1}{24} \left( \ell \sqrt{5} + m \right) - n \log_e \left( 1 + \sqrt{5} \right),\] where $ \ell, m, n \in \mathbb{N} $. Then $ \ell + m + n $ is equal to:

Let the area of the region $ \{(x, y): 0 \leq x \leq 3, 0 \leq y \leq \min\{x^2 + 2, 2x + 2\}\} $ be $ A $. Then $ 12A $ is equal to ______.

Area bounded by $0 ≤ y ≤ \text {min}(x^2+ 2, 2x + 2)$, $x∈[0, 3]$, then $12A$ is

$f(y - 2)^2 = (x - 1)$ and $x - 2y + 4 = 0$ then find the area bounded by the curves between the coordinate axis in first quadrant (in sq. units).

Area under the curve $x^2 + y^2 = 169$ and below the line $5x - y = 13$ is:

The probability distribution of a random variable is given below: \[\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X = x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X = x) & 0 & K & 2K & 2K & 3K & K^2 & 2K^2 & 7K^2 + K \\ \hline \end{array}\]

Find $ P(0<X<5) $.

Let $A$ be the area bounded by the curve $y=x|x-3|$, the $x$-axis and the ordinates $x=-1$ and $x=2$ Then $12 A$ is equal to _____

The area enclosed by the curves $y^2+4 x=4$ and $y-2 x=2$ is :

Let the area of the region $\left\{(x, y):|2 x-1| \leq y \leq\left|x^2-x\right|, 0 \leq x \leq 1\right\}$ be $A$ Then $(6 A +11)^2$ is equal to ____

The sum $1^2-2 \cdot 3^2+3 \cdot 5^2-4 \cdot 7^2+5 \cdot 9^2-\ldots+15 \cdot 29^2$ is_____

Area bounded by the curve 2y² = 3x and the line x+y = 3 outside the circle (x-3)² + y² = 2 and above the x-axis is A. The value of 4(π +4A) is?

If area bounded by region ${x, y)| |x^{2}-2| ≤ y ≤ x}$ is A, then 6A + 16$\sqrt{2}$ is?

Let the area of the region $\left\{(x, y):|2 x-1| \leq y \leq\left|x^2-x\right|, 0 \leq x \leq 1\right\}$ be $A$ Then $(6 A +11)^2$ is equal to ____