The area of the region bounded by the line $y = 4$ and the curve $y = x^2$ is ______.

(a) $\frac{32}{3}$ square units
(b) 0 square unit
(c) 1 square unit
(d) 32 square units
(e) $\frac{64}{3}$ square units
(f) $\frac{16}{3}$ square units
(g) 64 square units

Question

The area of the region bounded by the line $y = 4$ and the curve $y = x^2$ is ______.

(a) $\frac{32}{3}$ square units
(b) 0 square unit
(c) 1 square unit
(d) 32 square units
(e) $\frac{64}{3}$ square units
(f) $\frac{16}{3}$ square units
(g) 64 square units

Accepted Answer

>

> [!TLDR]
> Find the intersection points of $y=4$ and $y=x^2$, then integrate the vertical gap $(4 - x^2)$ between them.
>
> The curves intersect where $x^2 = 4$, giving $x = -2$ and $x = 2$. Between these limits the line $y=4$ lies above the parabola $y=x^2$, so the required area is
> $$A = \int_{-2}^{2}\big(4 - x^2\big)\,dx.$$
>
> Since the integrand is even, we can write $A = 2\int_0^2 (4 - x^2)\,dx$: …

Q.The area of the region bounded by the line $y = 4$ and the curve $y = x^2$ is ______.

Concept understanding — Area between Two Curves

Q.The area of the region bounded by the line y=4y = 4y=4 and the curve y=x2y = x^2y=x2 is ______.

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Q.The area of the region bounded by the line $y = 4$ and the curve $y = x^2$ is ______.

Unlock everything free for 14 days