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Area under the Curves

Section: Mathematics

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Question : 1 of 12

Marks: +1, -0

The area (in sq.units) of the region bounded by the curves

y=x^2

y=8-x^2

[AP EAPCET 24 May 2025 S1]

$\;\frac{32}{3}$
$\;\frac{16}{3}$
$\;\frac{64}{3}$
$\;\frac{128}{3}$

Solution:

Given, curves are

y=x^2

and

y=8-x^2

Equating these two equations, we get

\;x^2=8-x^2 \Rightarrow 2 x^2=8

\;\Rightarrow \;\; x^2=4 \Rightarrow x= \pm 2

\;\;\text{ Thus, area }\;=\int\limits_{-2}^{2} [ (8-x^2) -x^2] dx

\; \;\;\;\; \;=\int\limits_{-2}^{2} (8-x^2-x^2) dx=\int\limits_{-2}^{2} (8-2 x^2) dx

\;= \left[8 x-\;\frac{2}{3} x^3\right]_{-2}^2

\;= \{8(2)-\;\frac{2}{3}(2)^3\} - \{8(-2)-\;\frac{2}{3}(-2)^3\}

\;= \left(16-\;\frac{16}{3}\right) - \left(-16+\;\frac{16}{3}\right)

\;=16-\;\frac{16}{3}+16-\;\frac{16}{3}

\;=32-\;\frac{32}{3}=\;\frac{96-32}{3}=\;\frac{64}{3}

∴ The area of the region bounded by the curves is

\;\frac{64}{3}

sq. units.

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