If int_{0}^{1}(5x^{2}-3x+k)dx=0,then k= | vedclass

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(D) Given the integral equation: $\int_{0}^{1}(5x^{2}-3x+k)dx=0$ Integrating term by term: $\left[\frac{5x^{3}}{3} - \frac{3x^{2}}{2} + kx\right]_{0}^

Vedclass · Accepted Answer

$\frac{-1}{6}$

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(D) Given the integral equation: $\int_{0}^{1}(5x^{2}-3x+k)dx=0$ Integrating term by term: $\left[\frac{5x^{3}}{3} - \frac{3x^{2}}{2} + kx\right]_{0}^{1} = 0$ Applying the limits: $\left(\frac{5(1)^{3}}{3} - \frac{3(1)^{2}}{2} + k(1)\right) - (0) = 0$ Simplifying the expression: $\frac{5}{3} - \frac{3}{2} + k = 0$ Finding a common denominator: $\frac{10-9}{6} + k = 0$ $\frac{1}{6} + k = 0$ Therefore,$k = -\frac{1}{6}$

If $\int_{0}^{1}(5x^{2}-3x+k)dx=0$,then $k=$

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