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

Question

$A_1$ is the area bounded by $y=x^2+2$, $x+y=8$, and the $y$-axis in the first quadrant, and $A_2$ is the area bounded by $y=x^2+2$, $y^2=x$, $x=0$ and $x=2$ in the first quadrant. Find $(A_1-A_2)$.

Answer 1

Step 1: Computing the plain signed integral first:
$\displaystyle\int_{-1}^{1}(3x+2)dx=\Big[\tfrac{3x^2}{2}+2x\Big]_{-1}^{1}=\big(\tfrac32+2\big)-\big(\tfrac32-2\big)=\tfrac72-(-\tfrac12)=4$ — this is the NET signed area, not what's asked.

Step 2: Recognising why the signed value is wrong here:
Because the line dips below the x-axis for $x\in[-1,-2/3]$, that portion contributes negatively to the signed integral, undercounting the true (unsigned) area; we must instead treat the two pieces separately, as the region asked for is bounded by the axis regardless of sign.

Step 3: Recomputing with the correct split and absolute values:
Below-axis piece $x\in[-1,-2/3]$ contributes $|{-1/6}|=1/6$; above-axis piece $x\in[-2/3,1]$ contributes $25/6$; total $=26/6=13/3$.

Final Answer:
\[ \boxed{\dfrac{13}{3}\text{ sq. units}} \]

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

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Solution and Explanation

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