Find all values of k for which integral of x^2 from 2 to k =0 ?

To find all values of ( k ) for which ( \int_{2}^{k} x^2 , dx = 0 ), we need to evaluate the definite integral and set it equal to zero. The antiderivative of ( x^2 ) is ( \frac{1}{3}x^3 ). Applying the Fundamental Theorem of Calculus, we have:

[ \int_{2}^{k} x^2 , dx = \left[ \frac{1}{3}x^3 \right]_{2}^{k} ]

[ = \frac{1}{3}(k^3) - \frac{1}{3}(2^3) ]

[ = \frac{1}{3}k^3 - \frac{8}{3} ]

To find when this integral equals zero, we set it equal to zero:

[ \frac{1}{3}k^3 - \frac{8}{3} = 0 ]

[ \frac{1}{3}k^3 = \frac{8}{3} ]

Multiplying both sides by 3 to clear the fraction:

[ k^3 = 8 ]

Taking the cube root of both sides:

[ k = \sqrt[3]{8} ]

[ k = 2 ]

Therefore, the only value of ( k ) for which the integral of ( x^2 ) from 2 to ( k ) equals zero is ( k = 2 ).

To find the values of ( k ) for which the integral of ( x^2 ) from 2 to ( k ) equals 0, you need to evaluate the definite integral and set it equal to 0.

The definite integral of ( x^2 ) from 2 to ( k ) can be expressed as:

[ \int_{2}^{k} x^2 , dx ]

To find the antiderivative of ( x^2 ), integrate ( x^2 ) with respect to ( x ):

[ \int x^2 , dx = \frac{x^3}{3} + C ]

Now, evaluate the definite integral:

[ \left[ \frac{x^3}{3} \right]_{2}^{k} ]

[ = \frac{k^3}{3} - \frac{2^3}{3} ]

[ = \frac{k^3}{3} - \frac{8}{3} ]

Now, set the result equal to 0:

[ \frac{k^3}{3} - \frac{8}{3} = 0 ]

[ k^3 - 8 = 0 ]

Now, solve for ( k ):

[ k^3 = 8 ]

[ k = \sqrt[3]{8} ]

[ k = 2 ]

So, the value of ( k ) for which the integral of ( x^2 ) from 2 to ( k ) equals 0 is ( k = 2 ).