Question

Find the indefinite integral and check the result by differentiation. $$ \int\left(2 x-4^{x}\right) d x $$

Short answer

The indefinite integral of \(2x - 4^x\) is \(x^2 - \frac{1}{\ln 4} e^{x \ln 4} + C\), and it has been confirmed by differentiation.

Step-by-step solution

Finding the Integral

To find the indefinite integral of the given function \(2x - 4^x\), we break it into two integrals based on the subtraction operation. It becomes \(\int 2x \, dx\) and \(- \int 4^x \, dx\). We apply the integral calculus rule to the first part (which follows the power rule for integration) and we get \(x^2\). The second part is a bit tricky. We can't directly apply the exponential integration rule because the base is not 'e'. However, we can utilize a change of base formula that lets us rewrite this term as \(e^{x \ln 4}\). Then we integrate \(- \int e^{x \ln 4} \, dx\) using the exponential integration rule. The result is \(-\frac{1}{\ln 4} e^{x \ln 4}\). We do not forget to add the constant of integration 'C'. Hence the integral solution becomes \(x^2 - \frac{1}{\ln 4} e^{x \ln 4} + C\).

Checking the Result by Differentiation

In order to confirm our solution, we differentiate our result by using the power rule and chain rule. We first differentiate \(x^2\) and we get \(2x\). Then, we take the derivative of \(-\frac{1}{\ln 4} e^{x \ln 4}\) which becomes \(-4^x\). And the derivative of the constant 'C' is 0. Combining these gives us back the original function \(2x - 4^x\) which validates our solution.