Given that y=(x^3-2)^5 , find (dy)/(dx) b Hence , evaluate int _(0)^1x^2(x^3-2)^4dx __

To find the derivative of \( y = (x^3 - 2)^5 \), we can use the chain rule. Let \( u = x^3 - 2 \), then \( y = u^5 \).<br /><br />The derivative of \( y \) with respect to \( x \) is:<br />\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]<br /><br />First, compute \( \frac{dy}{du} \):<br />\[ \frac{dy}{du} = 5u^4 \]<br /><br />Next, compute \( \frac{du}{dx} \):<br />\[ \frac{du}{dx} = 3x^2 \]<br /><br />Now, combine these results:<br />\[ \frac{dy}{dx} = 5u^4 \cdot 3x^2 = 15x^2 (x^3 - 2)^4 \]<br /><br />So, the derivative is:<br />\[ \frac{dy}{dx} = 15x^2 (x^3 - 2)^4 \]<br /><br />Next, we need to evaluate the integral:<br />\[ \int_0^1 x^2 (x^3 - 2)^4 \, dx \]<br /><br />We can use the substitution method. Let \( u = x^3 - 2 \), then \( du = 3x^2 \, dx \), or \( x^2 \, dx = \frac{1}{3} \, du \).<br /><br />Change the limits of integration:<br />- When \( x = 0 \), \( u = -2 \)<br />- When \( x = 1 \), \( u = -1 \)<br /><br />Rewrite the integral in terms of \( u \):<br />\[ \int_0^1 x^2 (x^3 - 2)^4 \, dx = \int_{-2}^{-1} u^4 \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{-2}^{-1} u^4 \, du \]<br /><br />Evaluate the integral:<br />\[ \frac{1}{3} \int_{-2}^{-1} u^4 \, du = \frac{1}{3} \left[ \frac{u^5}{5} \right]_{-2}^{-1} = \frac{1}{15} \left[ u^5 \right]_{-2}^{-1} \]<br /><br />Calculate the definite integral:<br />\[ \frac{1}{15} \left[ (-1)^5 - (-2)^5 \right] = \frac{1}{15} \left[ -1 - (-32) \right] = \frac{1}{15} \left[ -1 + 32 \right] = \frac{1}{15} \cdot 31 = \frac{31}{15} \]<br /><br />So, the value of the integral is:<br />\[ \int_0^1 x^2 (x^3 - 2)^4 \, dx = \frac{31}{15} \]