We define 𝐶[𝑎, 𝑏] as the set of all continuous functions 𝑓:[𝑎, 𝑏] → ℝ. Let 𝑊 = ቄ𝑓 ∈ 𝐶[𝑎, 𝑏]: ∫ 𝑓(𝑥)𝑑𝑥 ௕ ௔ = 0ቅ. Show that 𝑊 is a subspace of 𝐶[𝑎, 𝑏].

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To show that 𝑊 is a subspace of 𝐶[𝑎, 𝑏], we need to prove three properties:1. 𝑊 is closed under addition.2. 𝑊 is closed under scalar multiplication.3. 𝑊 contains the zero function.Let&#039;s go through each of these properties:1. 𝑊 is closed under addition:To show that 𝑊 is closed under addition, we need to prove that if 𝑓₁, 𝑓₂ ∈ 𝑊, then 𝑓₁ + 𝑓₂ also belongs to 𝑊.Let 𝑓₁, 𝑓₂ ∈ 𝑊, which means ∫(𝑓₁(𝑥))𝑑𝑥 = 0 and ∫(𝑓₂(𝑥))𝑑𝑥 = 0.We want to show that ∫((𝑓₁ + 𝑓₂)(𝑥))𝑑𝑥 = 0.Using the linearity of the integral, we have:∫((𝑓₁ + 𝑓₂)(𝑥))𝑑𝑥 = ∫(𝑓₁(𝑥) + 𝑓₂(𝑥))𝑑𝑥.Now, let&#039;s compute this integral:∫(f1(x)+f2(x))dx=∫f1(x)dx+∫f2(x)dx.Since ∫(𝑓₁(𝑥))𝑑𝑥 = 0 and ∫(𝑓₂(𝑥))𝑑𝑥 = 0, we can substitute these values:∫(f1(x)+f2(x))dx=0+0=0.Therefore, 𝑓₁ + 𝑓₂ also belongs to 𝑊, and 𝑊 is closed under addition.2. 𝑊 is closed under scalar multiplication:To show that 𝑊 is closed under scalar multiplication, we need to prove that if 𝑓 ∈ 𝑊 and 𝑐 is a scalar, then 𝑐𝑓 also belongs to 𝑊.Let 𝑓 ∈ 𝑊, which means ∫(𝑓(𝑥))𝑑𝑥 = 0.We want to show that ∫((𝑐𝑓)(𝑥))𝑑𝑥 = 0.Using the linearity of the integral, we have:∫((𝑐𝑓)(𝑥))𝑑𝑥 = 𝑐∫(𝑓(𝑥))𝑑𝑥.Now, let&#039;s compute this integral:∫(c·f(x))dx=c·∫f(x)dx.Since ∫(𝑓(𝑥))𝑑𝑥 = 0, we can substitute this value:∫(c·f(x))dx=c·0=0.Therefore, 𝑐𝑓 also belongs to 𝑊, and 𝑊 is closed under scalar multiplication.3. 𝑊 contains the zero function:To show that 𝑊 contains the zero function, we need to prove that the zero function 𝑓(𝑥) = 0 belongs to 𝑊.To do this, we need to compute the integral:∫f(x)dx=∫0dx=0.Since the integral of the zero function over any interval is zero, the zero function 𝑓(𝑥) = 0 belongs to 𝑊.Therefore, 𝑊 satisfies all three properties: it is closed under addition, closed under scalar multiplication, and contains the zero function. Hence, 𝑊 is a subspace of 𝐶[𝑎, 𝑏].

We define 𝐶[𝑎, 𝑏] as the set of

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