Suppose the graph of $f$ consists of two regions between $x=0$ and $x=4$: from $x=0$ to $x=2$, $f\left(x\right)$ forms a triangle above the $x$-axis with area $3$; from $x=2$ to $x=4$, $f\left(x\right)$ forms a rectangle below the $x$-axis with area $4$. What is the value of the definite integral ${\int }_0^{4}f\left(x\right)dx$?

• A. $-7$
• B. $7$
• C. $-1$
• D. $1$

Answer: (C)

Verified step by step guidance

1. Step 1: Recall the definition of a definite integral. The definite integral \( \int_a^b f(x) \, dx \) represents the net area between the graph of \( f(x) \) and the x-axis over the interval \([a, b]\). Areas above the x-axis contribute positively, while areas below the x-axis contribute negatively.
2. Step 2: Analyze the given graph of \( f(x) \). From \( x = 0 \) to \( x = 2 \), \( f(x) \) forms a triangle above the x-axis with an area of 3. This contributes \( +3 \) to the integral.
3. Step 3: From \( x = 2 \) to \( x = 4 \), \( f(x) \) forms a rectangle below the x-axis with an area of 4. Since this region is below the x-axis, it contributes \( -4 \) to the integral.
4. Step 4: Combine the contributions from both regions. The net area is \( +3 - 4 \), which simplifies to \( -1 \).
5. Step 5: Conclude that the value of the definite integral \( \int_0^4 f(x) \, dx \) is \( -1 \). This result reflects the net area between the graph of \( f(x) \) and the x-axis over the interval \([0, 4]\).