Question

Find the area surrounded by the line \(y=3x+2\), the \(x\)-axis, and the ordinates \(x=-1\) and \(x=1\).

Hint:

Split the integral at the x-intercept x=-2/3 and take absolute values on the part below the axis.

Answer 1

Step 1: Finding where the line crosses the x-axis:
Set \(y=0\): \(3x+2=0\ \Rightarrow\ x=-\dfrac23\). This lies between \(-1\) and \(1\), so the line is below the x-axis on \(\big[-1,-\tfrac23\big]\) and above it on \(\big[-\tfrac23,1\big]\).

Step 2: Setting up the two separate integrals:
Area \(=\displaystyle\left|\int_{-1}^{-2/3}(3x+2)\,dx\right|+\int_{-2/3}^{1}(3x+2)\,dx\).

Step 3: Evaluating the antiderivative:
\(\displaystyle\int(3x+2)dx=\dfrac{3x^{2}}{2}+2x\).

Step 4: Computing the first piece (x=-1 to -2/3):
At \(x=-\tfrac23\): \(\dfrac{3(4/9)}{2}+2(-\tfrac23)=\tfrac{2}{3}-\tfrac{4}{3}=-\tfrac23\). At \(x=-1\): \(\dfrac{3}{2}-2=-\tfrac12\). Integral \(=-\tfrac23-(-\tfrac12)=-\tfrac16\); taking absolute value gives \(\tfrac16\).

Step 5: Computing the second piece (x=-2/3 to 1):
At \(x=1\): \(\dfrac32+2=\tfrac72\). At \(x=-\tfrac23\): \(-\tfrac23\) (from above). Integral \(=\tfrac72-(-\tfrac23)=\tfrac{21}{6}+\tfrac{4}{6}=\tfrac{25}{6}\).

Final Answer:
Total area \(=\tfrac16+\tfrac{25}{6}=\tfrac{26}{6}=\tfrac{13}{3}\) square units.\[ \boxed{\dfrac{13}{3}\text{ sq. units}} \]