Q. 32 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to write down three antiderivatives of each function f in Exercises 31–34.

$f\left(x\right)=\frac{1}{\sqrt{1+e^{4x}}}$

Step-by-Step Solution

Answer

The three antiderivatives for the function $f\left(x\right)=\frac{1}{\sqrt{1+e^{4x}}}$ are $F\left(x\right)={\int }_{-2}^x\frac{1}{\sqrt{1+e^{4t}}}dt, F\left(x\right)={\int }_{-1}^x\frac{1}{\sqrt{1+e^{4t}}}dt, F\left(x\right)={\int }_0^x\frac{1}{\sqrt{1+e^{4t}}}dt$.

Step 1. Given Information.

The function:
$f\left(x\right)=\frac{1}{\sqrt{1+e^{4x}}}$

Step 2. Graph the function.

Graph the function.

Step 3. Find the anti-derivatives.

If F is an anti-derivative of f, f is continuous on [a,b], then $F\left(x\right)={\int }_a^{x}f\left(t\right).dt$ for all $x\in [a,b]$.

So, $F\left(x\right)={\int }_{-2}^x\frac{1}{\sqrt{1+e^{4t}}}dt$ is defined to be an anti-derivative of the given function.

Similarly, the other two anti-derivatives are:

$F\left(x\right)={\int }_{-1}^x\frac{1}{\sqrt{1+e^{4t}}}dt, F\left(x\right)={\int }_0^x\frac{1}{\sqrt{1+e^{4t}}}dt$