integral from 0 to 2pi of sin(θ)

{ "query": { "display": "$$\\int_{0}^{2π}\\sin\\left(θ\\right)dθ$$", "symbolab_question": "BIG_OPERATOR#\\int _{0}^{2π}\\sin(θ)dθ" }, "solution": { "level": "PERFORMED", "subject": "Calculus", "topic": "Integrals", "subTopic": "Definite Integrals", "default": "0", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\int_{0}^{2π}\\sin\\left(θ\\right)dθ=0$$", "input": "\\int_{0}^{2π}\\sin\\left(θ\\right)dθ", "steps": [ { "type": "step", "primary": "Use the common integral: $$\\int\\:\\sin\\left(θ\\right)dθ=-\\cos\\left(θ\\right)$$", "result": "=[-\\cos\\left(θ\\right)]_{0}^{2π}" }, { "type": "interim", "title": "Compute the boundaries:$${\\quad}0$$", "input": "[-\\cos\\left(θ\\right)]_{0}^{2π}", "steps": [ { "type": "step", "primary": "$$\\int_{a}^{b}{f\\left(x\\right)dx}=F\\left(b\\right)-F\\left(a\\right)=\\lim_{x\\to\\:b-}\\left(F\\left(x\\right)\\right)-\\lim_{x\\to\\:a+}\\left(F\\left(x\\right)\\right)$$" }, { "type": "interim", "title": "$$\\lim_{θ\\to\\:0+}\\left(-\\cos\\left(θ\\right)\\right)=-1$$", "input": "\\lim_{θ\\to\\:0+}\\left(-\\cos\\left(θ\\right)\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$θ=0$$", "result": "=-\\cos\\left(0\\right)", "meta": { "title": { "extension": "Limit properties - if the limit of f(x), and g(x) exists, then:<br/>$$\\bullet\\quad\\lim_{x\\to\\:a}\\left(x\\right)=a$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[c\\cdot{f\\left(x\\right)}]=c\\cdot\\lim_{x\\to{a}}{f\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[\\left(f\\left(x\\right)\\right)^c]=\\left(\\lim_{x\\to{a}}{f\\left(x\\right)}\\right)^c$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\pm{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\pm\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\cdot{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\cdot\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}\\left(\\frac{f\\left(x\\right)}{g\\left(x\\right)}\\right)=\\frac{\\lim_{x\\to{a}}{f\\left(x\\right)}}{\\lim_{x\\to{a}}{g\\left(x\\right)}},\\:$$where $$\\lim_{x\\to{a}}g\\left(x\\right)\\neq0$$" } } }, { "type": "step", "primary": "Use the following trivial identity:$${\\quad}\\cos\\left(0\\right)=1$$", "secondary": [ "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$" ], "result": "=-1", "meta": { "solvingClass": "Solver" } } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "interim", "title": "$$\\lim_{θ\\to\\:2π-}\\left(-\\cos\\left(θ\\right)\\right)=-1$$", "input": "\\lim_{θ\\to\\:2π-}\\left(-\\cos\\left(θ\\right)\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$θ=2π$$", "result": "=-\\cos\\left(2π\\right)", "meta": { "title": { "extension": "Limit properties - if the limit of f(x), and g(x) exists, then:<br/>$$\\bullet\\quad\\lim_{x\\to\\:a}\\left(x\\right)=a$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[c\\cdot{f\\left(x\\right)}]=c\\cdot\\lim_{x\\to{a}}{f\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[\\left(f\\left(x\\right)\\right)^c]=\\left(\\lim_{x\\to{a}}{f\\left(x\\right)}\\right)^c$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\pm{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\pm\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\cdot{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\cdot\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}\\left(\\frac{f\\left(x\\right)}{g\\left(x\\right)}\\right)=\\frac{\\lim_{x\\to{a}}{f\\left(x\\right)}}{\\lim_{x\\to{a}}{g\\left(x\\right)}},\\:$$where $$\\lim_{x\\to{a}}g\\left(x\\right)\\neq0$$" } } }, { "type": "interim", "title": "$$\\cos\\left(2π\\right)=1$$", "input": "\\cos\\left(2π\\right)", "steps": [ { "type": "interim", "title": "$$\\cos\\left(2π\\right)=\\cos\\left(0\\right)$$", "input": "\\cos\\left(2π\\right)", "result": "=\\cos\\left(0\\right)", "steps": [ { "type": "step", "primary": "Rewrite $$2π$$ as $$2π+0$$", "result": "=\\cos\\left(2π+0\\right)" }, { "type": "step", "primary": "Apply the periodicity of $$\\cos$$: $$\\cos\\left(x+2π\\right)=\\cos\\left(x\\right)$$", "secondary": [ "$$\\cos\\left(2π+0\\right)=\\cos\\left(0\\right)$$" ], "result": "=\\cos\\left(0\\right)" } ], "meta": { "interimType": "N/A" } }, { "type": "interim", "title": "Use the following trivial identity:$${\\quad}\\cos\\left(0\\right)=1$$", "input": "\\cos\\left(0\\right)", "steps": [ { "type": "step", "primary": "$$\\cos\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\cos(x)&x&\\cos(x)\\\\\\hline 0&1&π&-1\\\\\\hline \\frac{π}{6}&\\frac{\\sqrt{3}}{2}&\\frac{7π}{6}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{1}{2}&\\frac{4π}{3}&-\\frac{1}{2}\\\\\\hline \\frac{π}{2}&0&\\frac{3π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{1}{2}&\\frac{5π}{3}&\\frac{1}{2}\\\\\\hline \\frac{3π}{4}&-\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&-\\frac{\\sqrt{3}}{2}&\\frac{11π}{6}&\\frac{\\sqrt{3}}{2}\\\\\\hline \\end{array}$$" }, { "type": "step", "result": "=1" } ], "meta": { "interimType": "Trig Trivial Angle Value Title 0Eq" } }, { "type": "step", "result": "=1" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7CJflLzwuxA99/WHYVkAK6CAn9lkDfZkicUGkO3EF+IrsnkTmAaLbHByGREyOhUnuiUqxJ9VFH79Cjtc7RciAvsWevzcXK72y5Ob8iTLU5Bs=" } }, { "type": "step", "result": "=-1" } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "step", "result": "=-1-\\left(-1\\right)" }, { "type": "step", "primary": "Simplify", "result": "=0", "meta": { "solvingClass": "Solver" } } ], "meta": { "interimType": "Integral Definite Limit Boundaries 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s79MtBbel9CwnBVqBrvUcE4yv2NQAyED8Wdm7cRryD+v8w4c+9R+KfwLcKhQIUqlKF6AVRxGB0xcze/HY0rc/oFsiU7OKsuJHQygybYV0fvWNeqXxdc+rps1CUyb7fqI2GYiwAciRdF7q/QJD6oWYm1bdPwo7Qm1U6oQSgO9j9pej" } }, { "type": "step", "result": "=0" } ], "meta": { "solvingClass": "Integrals" } }, "plot_output": { "meta": { "plotInfo": { "variable": "θ", "plotRequest": "yes" }, "showViewLarger": true } }, "meta": { "showVerify": true } }