Question

Find the indefinite integral and check your result by differentiation. $$ \int 4 y^{-3} d y $$

Short answer

The indefinite integral of \(4 y^{-3}\) is \(\frac{4}{y} + C\)

Step-by-step solution

Calculate the Integral

To solve the given integral, the power rule for integration will be applied. The first step is thus to add 1 to the exponent of \(y\), which is -3, and then divide the result by the new exponent. \n The integral \(\int 4 y^{-3} d y\) can be rewritten as \(-4 \int y^{-3 + 1}dy\), yielding \(-4 \int y^{-2} dy\). When applying the rule, the result is \(-4 \left(\frac{1}{-2 +1} y^{-2+1} \right) + C\). The \(+ C\) represents the constant of integration.

Simplify the Result

The next step is to simplify the result from the above step. The integral equals \(-4 \left(-1y^{-1} \right) + C = 4 y^{-1} + C\), which can also be written as \(\frac{4}{y} + C\).

Check the Result by Differentiation

Lastly, verify the result by differentiation. Taking the derivative of \(\frac{4}{y} + C\) with respect to \(y\) should yield the original function, \(4y^{-3}\). The derivative \(\frac{d}{dy}(\frac{4}{y} + C) = -4y^{-2 - 1} = -4y^{-3}\). After cancelling out the negative signs, it is concluded that the derivative of the result matches the original function, therefore confirming that the integral was computed correctly.