Consider the region bounded by $f (x) = x^{3}$ , the x-axis, $x = 0$ , and the line $x = 4$ . If the line $x = K$ divides this region into two parts of equal area, then the value of $K^{2}$ is

4

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4 \sqrt{2}

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8

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8 \sqrt{2}

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Solution

The correct option is D $8 \sqrt{2}$
$2 \int_{0}^{K} x^{3} d x = \int_{0}^{4} x^{3} d x$
$\frac{K^{4}}{2} = 64$
$K^{4} = 128$
$K^{2} = 8 \sqrt{2}$ .

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Consider the region bounded by f(x)=x3, the x-axis, x=0, and the line x=4. If the line x=K divides this region into two parts of equal area, then the value of K2 is

The correct option is D 8√22∫K0x3dx=∫40x3dxK42=64K4=128K2=8√2.

Consider the region bounded by $f (x) = x^{3}$ , the x-axis, $x = 0$ , and the line $x = 4$ . If the line $x = K$ divides this region into two parts of equal area, then the value of $K^{2}$ is

The correct option is D $8 \sqrt{2}$
$2 \int_{0}^{K} x^{3} d x = \int_{0}^{4} x^{3} d x$
$\frac{K^{4}}{2} = 64$
$K^{4} = 128$
$K^{2} = 8 \sqrt{2}$ .