Find the antiderivative of e^{-0.5t} + 1


Find the antiderivative of {eq}e^{-0.5t} + 1 {/eq}


The antiderivative of any function is also known as the indefinite integral of the function.

To solve the indefinite integral, we'll use the integral sum rule, which is written as {eq}\int (f(x)+ g(x))dx=\int f(x)dx + \int g(x)dx {/eq}

Next, we use the common integral formulas:

{eq}\int e^{ax}dx=\dfrac{e^{ax}}{a}\\ \int dx=x {/eq}

Answer and Explanation:

We need to find out the antiderivative of {eq}e^{-0.5t} + 1 {/eq}

Hence, we need to compute {eq}\int (e^{-0.5t} + 1) dt {/eq}


{eq}\begin{align} \int (e^{-0.5t} + 1) dt &=\int e^{-0.5t}dt+\int 1dt \ & \left [ \because \int (f(x)+ g(x))dx=\int f(x)dx + \int g(x)dx \right ]\\ &=\dfrac{e^{-0.5t}}{-0.5}+t+C \ & \left [ \because \int e^{ax}dx=\dfrac{e^{ax}}{a}, \int dx=x \right ]\\ &=-\dfrac{e^{-0.5t}}{0.5}+t+C\\ &=-\dfrac{e^{-0.5t}}{\frac{5}{10}}+t+C\\ &=-2e^{-0.5t}+t+C \end{align} {/eq}

where {eq}C {/eq} is constant of integration.

{eq}\color{blue}{\boxed{\int (e^{-0.5t} + 1) dt=-2e^{-0.5t}+t+C}} {/eq}

Learn more about this topic:

Antiderivative: Rules, Formula & Examples

from Calculus: Help and Review

Chapter 8 / Lesson 12

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