Q.Find the area of ellipse $\dfrac{x^2}{1} + \dfrac{y^2}{4} = 1$.

The parabola $y^2=4x$ has $a=1$; the latus rectum is $x=1$. Area $=\tfrac{8}{3}$ sq.units.

Compare $y^2=4x$ with the standard form $y^2=4ax$: $4a=4\Rightarrow a=1$. The latus rectum is the vertical line $x=a=1$. The region is bounded by the parabola and this line, symmetric about the $x$-axis, so

$A = 2\int_0^{1} y\,dx = 2\int_0^{1} 2\sqrt{x}\,dx = 4\int_0^{1} x^{1/2}\,dx.$

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For $x\ge0$, $|x|=x$; the bounded area is $\int_0^2 x\,dx=2$ square units, option (a).

Simplify the curve. For $0\le x\le 2$, $|x|=x$, so $y=x$.

Area under the curve.

$A=\int_0^2 x\,dx=\left[\frac{x^2}{2}\right]_0^2=\frac{4}{2}-0=2.$

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On $[0,2]$, $y=|x|=x$, so the area is $\int_0^2 x\,dx=2$ sq. units.

Between $x=0$ and $x=2$ the graph $y=|x|$ coincides with the line $y=x$ through the origin, which lies above the $x$-axis. The bounded area is

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Area $=\int_0^1 e^x\,dx=[e^x]_0^1=e-1$ sq. units.

Since $e^x>0$ throughout $[0,1]$, the region between the curve and the $x$-axis has area

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The required area is $\displaystyle\int_0^3 x^2\, dx = \left[\dfrac{x^3}{3}\right]_0^3 = 9$ square units.

The region is bounded above by the curve $y = x^2$, below by the X-axis, and between the lines $x = 0$ and $x = 3$. Since $y = x^2 \ge 0$ throughout, the area equals

$A = \int_0^3 y\, dx = \int_0^3 x^2\, dx.$

Evaluating: …

Between $0$ and $2$, $|x| = x$, so the required area is $\int_0^2 x\,dx = \left[\frac{x^2}{2}\right]_0^2 = 2$ sq.units.

Step 1 — Simplify $|x|$ on the interval. For $x \in [0, 2]$, $x \ge 0$, hence $y = |x| = x$.

Step 2 — Set up and evaluate the definite integral. …