Answer
See below
Work Step by Step
Given: $V=C[a,b]\\
S=\{f \in V:\int^b_a f(x)dx=0\}$
Take $f \in C[a,b]$ such that $\int^b_a f(x)dx=0 \forall x \in [a,b]\\
\rightarrow f \in W$
We can see $W$ is nonempty (1)
Let $f,g \in W \\\int^b_a f(x)dx=0\\
\int^b_a g(x)dx=0$
then $\int^b_a (f+g)(x)dx=\int^b_a (f(x)+g(x))dx=\int^b_a f(x)dx+\int^b_a g(x)dx=0\\
\rightarrow f+g \in C[a,b]\\
\rightarrow f+g \in W$
Hence, $f+g1$ is closed under addition multiplication (2)
Let $k$ be a scalar
Obtain $\int^b_a (kf)(x)dx=\int^b_a kf(x)dx=k\int^b_a f(x)dx=k.0=0\\
\rightarrow kf \in C[a,b]\\
\rightarrow kf \in W$
Hence, $kf$ is closed under scalar multiplication (3)
From (1)(2)(3), $W$ is a subspace of $C[a,b]$
