Question

the area of the region bounded between the given graph of $f$ and the $x$-axis on $-3, 0$ is $16$, the area bounded between the graph of $f$ and the $x$-axis on $0, 2$ is $5$ and the area bounded between the graph of $f$ and the $x$-axis on $2, 4$ is $43$, use the following graph to evaluate $int_{-3}^{4} |f(x)| dx$.

Explanation:

Step1: Recall the property of absolute value integral

The integral of \(|f(x)|\) over an interval is the sum of the areas between the graph of \(f(x)\) and the \(x\)-axis on that interval, because \(|f(x)|\) makes all the areas positive (regardless of whether \(f(x)\) is above or below the \(x\)-axis). So we can split the integral \(\int_{-3}^{4}|f(x)|dx\) into three parts: \(\int_{-3}^{0}|f(x)|dx\), \(\int_{0}^{2}|f(x)|dx\) and \(\int_{2}^{4}|f(x)|dx\).

Step2: Evaluate each part

• For \(\int_{-3}^{0}|f(x)|dx\): The area between \(f(x)\) and the \(x\)-axis on \([-3,0]\) is given as 16. Since \(|f(x)|\) will make the integral equal to this area (because the area is already a positive quantity representing the region between the curve and the axis), so \(\int_{-3}^{0}|f(x)|dx = 16\).
• For \(\int_{0}^{2}|f(x)|dx\): The area between \(f(x)\) and the \(x\)-axis on \([0,2]\) is 5. So \(\int_{0}^{2}|f(x)|dx=5\) (again, because the area is positive and \(|f(x)|\) captures the magnitude of the area).
• For \(\int_{2}^{4}|f(x)|dx\): The area between \(f(x)\) and the \(x\)-axis on \([2,4]\) is 43. So \(\int_{2}^{4}|f(x)|dx = 43\).

Step3: Sum the parts

Now we sum these three integrals: \(\int_{-3}^{4}|f(x)|dx=\int_{-3}^{0}|f(x)|dx+\int_{0}^{2}|f(x)|dx+\int_{2}^{4}|f(x)|dx\)

Substitute the values we found: \(16 + 5+43\)

Step4: Calculate the sum

\(16+5 = 21\), then \(21 + 43=64\)

Answer:

\(64\)