If f(x) is a continuous,increasing,and odd fu | vedclass

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

Question

(A) Given that $f(x)$ is an odd function,we have $f(-x) = -f(x)$.We are given $\int_{-1}^{4} f(x) \,dx = 10$.This can be split as $\int_{-1}^{0} f(x)

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(A) Given that $f(x)$ is an odd function,we have $f(-x) = -f(x)$.We are given $\int_{-1}^{4} f(x) \,dx = 10$.This can be split as $\int_{-1}^{0} f(x) \,dx + \int_{0}^{4} f(x) \,dx = 10$.Since $f(x)$ is odd,$\int_{-1}^{0} f(x) \,dx = -\int_{0}^{1} f(x) \,dx = -\frac{3}{2}$.Substituting this,we get $-\frac{3}{2} + \int_{0}^{4} f(x) \,dx = 10$,which implies $\int_{0}^{4} f(x) \,dx = 10 + \frac{3}{2} = \frac{23}{2}$.The required area is $\int_{-4}^{4} |f(x)| \,dx$.Since $f(x)$ is an odd function,$|f(x)|$ is an even function,so the area is $2 \int_{0}^{4} |f(x)| \,dx$.Since $f(x)$ is increasing and $f(0) = 0$ (as it is odd),$f(x) \leq 0$ for $x \in [-4, 0]$ and $f(x) \geq 0$ for $x \in [0, 4]$.Thus,$\int_{0}^{4} |f(x)| \,dx = \int_{0}^{4} f(x) \,dx = \frac{23}{2}$.Therefore,the total area is $2 	imes \frac{23}{2} = 23$.

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is:

Similar Questions

The area of the smaller portion between the curves $x^2 + y^2 = 8$ and $y^2 = 2x$ is

The area enclosed by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$,and $y = |\ln |x||$ is equal to:

The area included between the two curves $y^2 = 4ax$ and $x^2 = 4ay$ is

If the area of the region $\{(x, y): |x^2-2| \leq y \leq x\}$ is $A$,then $6A + 16\sqrt{2}$ is equal to $...........$.

The area (in sq units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module