The value of $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$ is:

We need to find $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$.

The limits are symmetric around zero, suggesting special symmetry properties.

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Substitute $-x$ into the function:

$f(-x) = \frac{7^{-x}}{1+7^{-x}}$

Multiply both numerator and denominator by $7^x$:

$f(-x) = \frac{7^{-x}}{1+7^{-x}} \times \frac{7^x}{7^x}$

$f(-x) = \frac{7^{-x} \cdot 7^x}{(1+7^{-x}) \cdot 7^x}$

$f(-x) = \frac{1}{7^x + 1}$

$f(-x) = \frac{1}{1+7^x}$

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Add $f(x)$ and $f(-x)$:

$f(x) + f(-x) = \frac{7^x}{1+7^x} + \frac{1}{1+7^x}$

$f(x) + f(-x) = \frac{7^x + 1}{1+7^x}$

$f(x) + f(-x) = \frac{1+7^x}{1+7^x}$

$f(x) + f(-x) = 1$

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Let $I = \int_{-a}^a f(x)dx$

Replace $x$ with $-x$ using substitution. With symmetric limits, the integral remains unchanged:

$I = \int_{-a}^a f(-x)dx$

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Add the two forms:

$I + I = \int_{-a}^a f(x)dx + \int_{-a}^a f(-x)dx$

$2I = \int_{-a}^a [f(x) + f(-x)]dx$

$2I = \int_{-a}^a 1 \, dx$

$2I = x\bigg|_{-a}^{a}$

$2I = a - (-a)$

$2I = 2a$

$I = a$

Therefore, $\int_{-a}^a f(x)dx = a$