Consider the function f(x) = x^3 - 8x^2 + 20x | vedclass

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

Question

(A) The function is $f(x) = x^3 - 8x^2 + 20x - 13$. To find the area enclosed by the curve and the coordinate axes,we look at the region bounded by $x

Vedclass · Accepted Answer

$\frac{65}{12}$

Vedclass · Answer

(A) The function is $f(x) = x^3 - 8x^2 + 20x - 13$. To find the area enclosed by the curve and the coordinate axes,we look at the region bounded by $x=0$,$y=0$,and the curve $y=f(x)$. From the graph,the curve intersects the $x$-axis at $x=1$. The area $A$ is the absolute value of the integral of $f(x)$ from $x=0$ to $x=1$: $A = \left| \int_{0}^{1} (x^3 - 8x^2 + 20x - 13) \, dx \right|$ $A = \left| \left[ \frac{x^4}{4} - \frac{8x^3}{3} + 10x^2 - 13x \right]_{0}^{1} \right|$ $A = \left| \left( \frac{1}{4} - \frac{8}{3} + 10 - 13 \right) - 0 \right|$ $A = \left| \frac{3 - 32 + 120 - 156}{12} \right|$ $A = \left| \frac{-65}{12} \right| = \frac{65}{12}$ Thus,the area is $\frac{65}{12}$ square units.

Consider the function $f(x) = x^3 - 8x^2 + 20x - 13$. The area enclosed by $y = f(x)$ and the coordinate axes is:

Similar Questions

Find the area of the region bounded by the line $y=3x+2$,the $x$-axis,and the ordinates $x=-1$ and $x=1$.

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

If the ordinate $x = a$ divides the area bounded by the curve $y = \left( 1 + \frac{8}{x^2} \right)$,the $x$-axis,and the ordinates $x = 2$ and $x = 4$ into two equal parts,then $a = $

The area of the region bounded by the curve $y = x^3$,and the lines $y = 8$ and $x = 0$ is

The area of the region bounded by the parabola $y = x^2 + 2$,the $X$-axis,and the lines $x = 1$ and $x = 2$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module