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Question: What is the antiderivative of \({e^{2x}}\)?...

What is the antiderivative of e2x{e^{2x}}?

Explanation

Solution

Antiderivative is the opposite of derivative. It is the same as integrals or integration. In order to find the antiderivative of e2x{e^{2x}}, substitute the value of 2x=u2x = u in e2x{e^{2x}},then integrate according to the best suitable methods of integration known to us. And the best method that would be suitable for this function would be integration by substitution, that is replacing a function with a new variable, then integrating the function according to that variable only.

Formula used:
eydy=ey+C\int {{e^y}dy = {e^y} + C}

Complete step by step answer:
The antiderivative of the exponential e2x{e^{2x}}.
Substitute 2x=u2x = u in the exponential e2x{e^{2x}}, and we get: eu{e^u}.
Differentiating 2x=u2x = u with respect to xx, we get:
2x=u d2xdx=dudx 2dxdx=dudx 2dx=du dx=du2 2x = u \\\ \Rightarrow \dfrac{{d2x}}{{dx}} = \dfrac{{du}}{{dx}} \\\ \Rightarrow 2\dfrac{{dx}}{{dx}} = \dfrac{{du}}{{dx}} \\\ \Rightarrow 2dx = du \\\ \Rightarrow dx = \dfrac{{du}}{2} \\\
Integrating e2x{e^{2x}} with respect to dxdx;
e2xdx\int {{e^{2x}}} dx
Substituting the value of dx=du2dx = \dfrac{{du}}{2} and 2x=u2x = u in the exponential e2x{e^{2x}} in the above function, and we get:
eudu2\int {{e^u}} \dfrac{{du}}{2}
As, we know that eydy=ey+C\int {{e^y}dy = {e^y} + C}
So using this formula integrating the above function, we get:
12eudu=eu2+C\dfrac{1}{2}\int {{e^u}} du = \dfrac{{{e^u}}}{2} + C
Now, substituting back the original values of u=2xu = 2x in the above resultant and we get:
e2x2+C\dfrac{{{e^{2x}}}}{2} + C. We can also cross check the result by differentiating the output with respect to xx.

Therefore, the antiderivative of e2x{e^{2x}} is e2x2+C\dfrac{{{e^{2x}}}}{2} + C.

Note: It’s very important to add a constant after integrating the equation because when we derive an equation their constant is removed. For example, let’s see an equation: 2x+22x + 2. Derivative this with respect to xx and we get: d(2x+2)dx=d2xdx+d2dx=2+0=2\dfrac{{d(2x + 2)}}{{dx}} = \dfrac{{d2x}}{{dx}} + \dfrac{{d2}}{{dx}} = 2 + 0 = 2.Because we know that derivation of constant term is zero. Now Let’s check and integrate the answer obtained and we get: 2dx=2x\int {2dx} = 2x and we can see that the constant term is lost. That’s why we need to add a constant term at last after integration.