[amp_mcq option1=”$$ – 2\\pi $$” option2=”$$ – \\pi $$” option3=”$$\\pi $$” option4=”$$2\\pi $$” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\pi}$.<\/p>\n
To evaluate the integral, we can use integration by parts. Let $u=x^2$ and $dv=\\cos x \\,dx$. Then $du=2x \\,dx$ and $v=\\sin x$. Substituting into the formula for integration by parts, we get<\/p>\n
$$\\int\\limits_0^\\pi {{{\\text{x}}^2}\\cos \\,{\\text{x dx}}} =x^2 \\sin x \\bigg|_0^\\pi -\\int\\limits_0^\\pi 2x \\sin x \\,dx$$<\/p>\n
The first term on the right-hand side is $0$, since $x^2 \\sin x$ is an odd function and the limits of integration are $0$ and $\\pi$. The second term on the right-hand side can be evaluated using integration by parts again, with $u=x$ and $dv=\\sin x \\,dx$. This gives<\/p>\n
$$\\int\\limits_0^\\pi 2x \\sin x \\,dx =x\\cos x \\bigg|_0^\\pi -\\int\\limits_0^\\pi \\cos x \\,dx$$<\/p>\n
The first term on the right-hand side is $0$, since $x\\cos x$ is an odd function and the limits of integration are $0$ and $\\pi$. The second term on the right-hand side is $-\\sin x$, so the integral is equal to<\/p>\n
$$\\int\\limits_0^\\pi {{{\\text{x}}^2}\\cos \\,{\\text{x dx}}} =0-(-\\sin x) =\\pi$$<\/p>\n
Therefore, the value of the integral is $\\boxed{\\pi}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$ – 2\\pi $$” option2=”$$ – \\pi $$” option3=”$$\\pi $$” option4=”$$2\\pi $$” correct=”option1″]<\/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-20272","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n