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

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(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)

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$23$

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(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:

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