{"id":4000,"date":"2021-08-13T07:22:55","date_gmt":"2021-08-13T07:22:55","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4000"},"modified":"2021-11-21T01:11:22","modified_gmt":"2021-11-20T19:41:22","slug":"formulas-for-definite-integrals","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/formulas-for-definite-integrals\/","title":{"rendered":"Properties and Formulas for Definite Integrals"},"content":{"rendered":"
Here, you will learn formulas for definite integrals and properties of definite integrals with examples.<\/p>\n
Let’s begin –<\/p>\n
A definite integral is denoted by \\(\\int_{a}^{b}\\) f(x)dx which represent the algebraic area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis.<\/p>\n
\n(a) \\(\\int_{a}^{b}\\) f(x)dx = \\(\\int_{a}^{b}\\) f(t)dt provided f is same<\/p>\n
(b) \\(\\int_{a}^{b}\\) f(x)dx = – \\(\\int_{b}^{a}\\) f(x)dx<\/p>\n
(c) \\(\\int_{a}^{b}\\) f(x)dx = \\(\\int_{a}^{c}\\) f(x)dx + \\(\\int_{c}^{b}\\) f(x)dx , where c may lie inside or outside the interval [a,b]. This property is to be used when f is piecewise continous in (a, b).<\/p>\n
(d) \\(\\int_{a}^{a}\\) f(x)dx = \\(\\int_{0}^{a}\\) [f(x) + f(-x)]dx = \\(\\begin{cases} 0 & \\text{if f(x) is an odd function}\\ \\\\ 2\\int_{a}^{b} f(x)dx & \\text{if f(x) is an even function}\\ \\end{cases}\\)<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Evaluate \\(\\int_{1\/2}^{1\/2}\\) \\(cosx ln{({1+x\\over 1-x})}\\) dx<\/p>\n
Solution : <\/span>f(-x) = \\(cos(-x) ln{({1-x\\over 1+x})}\\) = – \\(cosx ln{({1+x\\over 1-x})}\\) = f(-x)
\n \\(\\implies\\) f(x) is odd
\n Hence, the value of the given interval is 0.
<\/p>\n\n\n\n(e) \\(\\int_{a}^{b}\\) f(x)dx = \\(\\int_{a}^{b}\\) f(a+b-x)dx, In particular \\(\\int_{0}^{a}\\) f(x)dx = \\(\\int_{0}^{a}\\) f(a-x)dx<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Evaluate \\(\\int_{0}^{\\pi\/2}\\) \\(asinx+bcosx\\over sinx+cosx\\) dx<\/p>\n
Solution : <\/span>I = \\(\\int_{0}^{\\pi\/2}\\) \\(asinx+bcosx\\over sinx+cosx\\) dx ….(i)
\n I = \\(\\int_{0}^{\\pi\/2}\\) \\(asin(\\pi\/2-x)+bcos(\\pi\/2-x)\\over sin(\\pi\/2-x)+cos(\\pi\/2-x)\\) dx = \\(\\int_{0}^{\\pi\/2}\\) \\(acosx+bsinx\\over sinx+cosx\\) dx ….(ii)
\n Adding (i) and (ii),
\n 2I = \\(\\int_{0}^{\\pi\/2}\\) \\(a+b)(sinx+cosx)\\over sinx+cosx\\) dx = \\(\\int_{0}^{\\pi\/2}\\) (a+b) dx = (a+b)\\(\\pi\/2\\)
\n \\(\\implies\\) I = (a+b)\\(\\pi\/4\\)
<\/p>\n\n\n\n(f) \\(\\int_{0}^{2a}\\) f(x)dx = \\(\\int_{0}^{a}\\) f(x)dx + \\(\\int_{0}^{a}\\) f(2a-x)dx = \\(\\begin{cases} 2\\int_{0}^{a} f(x)dx & \\text{if}\\ f(2a-x) = f(x) \\\\ 0 & \\text{if}\\ f(2a-x) = -f(x) \\end{cases}\\).<\/p>\n
(g) \\(\\int_{0}^{nT}\\) f(x)dx = n\\(\\int_{0}^{T}\\) f(x)dx, (n \\(\\in\\) I); where T is the period of the function i.e. f(T+x) = f(x)<\/p>\n
(h) \\(\\int_{a+nT}^{b+nT}\\) f(x)dx = \\(\\int_{a}^{b}\\) f(x)dx, where f(x) is periodic with period T & n \\(\\in\\) I.<\/p>\n
(i) \\(\\int_{mT}^{nT}\\) f(x)dx = (n-m)\\(\\int_{0}^{T}\\) f(x)dx, where f(x) is periodic with period T & (n, m \\(\\in\\) I).<\/p>\n<\/blockquote>\n\n\n