(A) The area $A$ bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by $A = \int_a^b y \, dx$. Given the curve $x

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(A) The area $A$ bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by $A = \int_a^b y \, dx$. Given the curve $xy = c$,we have $y = \frac{c}{x}$. Substituting the limits $x = 1$ and $x = 4$,the area is: $A = \int_1^4 \frac{c}{x} \, dx$ $A = c [\ln |x|]_1^4$ $A = c (\ln 4 - \ln 1)$ Since $\ln 1 = 0$ and $\ln 4 = \ln(2^2) = 2 \ln 2$,we get: $A = c(2 \ln 2) = 2c \ln 2 \text{ sq. units}$.

Class 12 Mathematics Application of Integration Area bounded by region of single curve

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The area bounded by the curve $xy = c$,the $x$-axis,and the lines $x = 1$ and $x = 4$ is:

A
$2c \log 2 \text{ sq. units}$
B
$2 \log c \text{ sq. units}$
C
$c \log 3 \text{ sq. units}$
D
$2c \log 5 \text{ sq. units}$

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Application of Integration

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Area bounded by region of single curve

$A$ curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$. The area bounded by the curve and the $x$-axis is:

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Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals:

DifficultJEE MAIN 2019

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The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \frac{3\pi}{2}$ is:

EasyGSEB 2024

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Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

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Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

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The area bounded by the curve $xy = c$,the $x$-axis,and the lines $x = 1$ and $x = 4$ is:

Similar Questions

$A$ curve satisfying the initial condition $y(1)= 0$ satisfies the differential equation $x \frac{dy}{dx}= y -x^2$. The area bounded by the curve and the $x$-axis is:

Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals:

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \frac{3\pi}{2}$ is:

Find the area of the region bounded by the curve $y=x^{2}$ and the line $y=4$. (in $/3$)

Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

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