[amp_mcq option1=”\\[\\frac{1}{6}\\]” option2=”\\[\\frac{1}{4}\\]” option3=”\\[\\frac{1}{3}\\]” option4=”\\[\\frac{1}{2}\\]” correct=”option3″]<\/p>\n
The area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x$-$y$ plane is $\\frac{1}{3}$.<\/p>\n
To find the area, we can use the following formula:<\/p>\n
$$\\text{Area}=\\int_{a}^{b}\\left(f(x)-g(x)\\right)\\,dx$$<\/p>\n
where $f(x)$ and $g(x)$ are the functions that define the boundaries of the area, and $a$ and $b$ are the $x$-coordinates of the endpoints of the area.<\/p>\n
In this case, $f(x)=x$ and $g(x)=x^2$, so we have:<\/p>\n
$$\\text{Area}=\\int_{0}^{1}\\left(x-x^2\\right)\\,dx=\\int_{0}^{1}-x^2\\,dx$$<\/p>\n
We can evaluate this integral using the following formula:<\/p>\n
$$\\int x^n\\,dx=\\frac{x^{n+1}}{n+1}+C$$<\/p>\n
where $C$ is an arbitrary constant.<\/p>\n
In this case, $n=2$, so we have:<\/p>\n
$$\\begin{align} Therefore, the area enclosed between the straight line $y=x$ and the parabola $y=x^2$ in the $x$-$y$ plane is $\\frac{1}{3}$.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”\\[\\frac{1}{6}\\]” option2=”\\[\\frac{1}{4}\\]” option3=”\\[\\frac{1}{3}\\]” option4=”\\[\\frac{1}{2}\\]” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20186","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n\\text{Area}&=\\int_{0}^{1}-x^2\\,dx \\\\
\n&=-\\frac{x^3}{3}~\\Bigg|_{0}^{1} \\\\
\n&=-\\frac{1}{3}-\\left(0\\right) \\\\
\n&=\\frac{1}{3}
\n\\end{align<\/em>}$$<\/p>\n