Let y = g(x) be the inverse of a bijective ma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = 3x^3 + 2x$. We need to find the area bounded by $g(x)$,the $x-$axis,and the line $x = 5$. Since $g(x)$ is the inverse of $f(x)$,the

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(D) Given $f(x) = 3x^3 + 2x$. We need to find the area bounded by $g(x)$,the $x-$axis,and the line $x = 5$. Since $g(x)$ is the inverse of $f(x)$,the area bounded by $g(x)$ from $x=0$ to $x=5$ is equivalent to the area bounded by $f(x)$ with the $y-$axis from $y=0$ to $y=5$. First,find the value of $x$ for which $f(x) = 5$: $3x^3 + 2x = 5 \implies 3x^3 + 2x - 5 = 0$. By inspection,$x = 1$ is a root because $3(1)^3 + 2(1) - 5 = 0$. The area is given by $\int_{0}^{5} g(x) \, dx$. Using the property of inverse functions,$\int_{0}^{a} g(x) \, dx + \int_{0}^{g(a)} f(x) \, dx = a \cdot g(a)$. Here $a = 5$ and $g(5) = 1$. So,$\int_{0}^{5} g(x) \, dx + \int_{0}^{1} (3x^3 + 2x) \, dx = 5 \cdot 1 = 5$. $\int_{0}^{1} (3x^3 + 2x) \, dx = [\frac{3x^4}{4} + x^2]_{0}^{1} = \frac{3}{4} + 1 = \frac{7}{4}$. Therefore,the required area is $5 - \frac{7}{4} = \frac{20 - 7}{4} = \frac{13}{4}$.

Let $y = g(x)$ be the inverse of a bijective mapping $f : R \rightarrow R$ defined by $f(x) = 3x^3 + 2x$. The area bounded by the graph of $g(x)$,the $x-$axis,and the ordinate at $x = 5$ is:

Similar Questions

The area of the region bounded by the curve $x + 2y + 8 = 0$ and the lines $y = -3$ and $y = -1$ is . . . . . . sq. units.

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 $12A$ is equal to $...........$.

If the area of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$ sq units,then $k$ equals

Area enclosed by the ellipse $2x^2 + 3y^2 = 1$ is . . . . . . sq. units.

The area of the region bounded by the curve $y = \tan x$,$x = 0$,$x = \frac{\pi}{4}$,and the $X$-axis is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module