Let y=g(x) be the inverse of a bijective mapping f: R arrow R f(x)=3 x3+2 x. The area bounded by the graph of g(x), the x-axis and the ordinate at x=5 is :

Question

Let y=g(x) be the inverse of a bijective mapping f: R arrow R f(x)=3 x3+2 x. The area bounded by the graph of g(x), the x-axis and the ordinate at x=5 is : (A) (5/4) (B) (7/4) (C) (9/4) (D) (13/4)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/de210f79deb2f2f081e63cfbf9e6ce4d-.png" /> note for inverse function y axis will be the x axis and x axis will be the y axis text required area = text Area of rectangle -∫ limits01 f(x) d x =5-∫ limits01(3 x3+2 x) d x =5-((3/4)+1)=3 (1/4)=(13/4)

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

$\frac{5}{4}$

$\frac{7}{4}$

$\frac{9}{4}$

$\frac{13}{4}$

note for inverse function $y$ axis will be the $x$ axis and $x$ axis will be the $y$ axis
$required area = Area of rectangle - 0 \int 1 f (x) d x$
$= 5 - 0 \int 1 (3 x^{3} + 2 x) d x$
$= 5 - (\frac{3}{4} + 1) = 3 \frac{1}{4} = \frac{13}{4}$

Q. Let y=g(x) be the inverse of a bijective mapping f:R→Rf(x)=3x3+2x. The area bounded by the graph of g(x), the x-axis and the ordinate at x=5 is :

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