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I attempted to solve this problem in the following manner:

Part1:

The area under the curve is defined by the integral below:

$$ \int_0^{10} g(x)dx $$

and can be easily obtained by using the Area of a triangle function

$$ \frac12bh $$

which in this case would equal to

$$ \frac1210*20 = 100$$

therefore: $$ \int_0^{10} g(x)dx = 100 $$

Part2 is another integral that is defined by the Area of a circle such that: $$ A = \pi r^2 $$

or since it is only a Half-Circle:

$$ A = \int_{10}^{30} g(x)dx = \pi \frac {r^2}2 = 50\pi $$

Before we can complete the 3rd part of the question you have to find:

$$ \int_{30}^{35} g(x)dx $$

using the same concept as in part1, the following is also true here: $$ \frac12bh $$

therefore:

$$ A = \int_{30}^{35} g(x)dx = \frac12 {5} * {5} = 12.5 $$

Now you can add it all together to say that:

$$ A = \int_{0}^{35} g(x)dx = 112.5 + 50\pi$$

however the above answer is incorrect. How can I solve this correctly based off the answer I just walked through?

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    The integral $\int\limits_{10}^{30} g(x)\, dx $ will be negative. – 2017-01-12
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    Remember that you can only write $A=\int_{10}^{30}g(x)dx$ if $g(x)$ is nonnegative for all $x\in[10,30]$. But in your graph, $-10\leq g(x)\leq 0$ for all $x\in[10,30]$. – 2017-01-12

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Your integrals are not all correct. Your first $2$ answers are correct, considering only the absolute values of the integrals. For the second and final one, observe that you have to use the concept of positive and negative areas, crudely speaking.

Note that the second integral is negative since the semi circle lies below the x axis.

So your second answer should actually be $\color{red}{-}50\pi$ and the final answer should be $112.5\color{red}{-}50\pi$.

Hope this helps you.