A' fuasgladh cho-aontaran diofarail

Thug sinn iomradh roimhe air an teirm \(+ c\). Bheir sinn a-nis sùil air mar a lorgas sinn luach \(c\) nuair a tha fiosrachadh a bharrachd air a thoirt dhuinn sa cho-aontar.

Eisimpleir 1 (leudachadh)

Obraich a-mach co-aontar na lùib far a bheil \(\frac{{dy}}{{dx}} = 4{x^3} + 6{x^2}\) agus a' dol tron phuing \((1,3)\).

Fuasgladh

\[y = \int {(4{x^3}} + 6{x^2})dx = {x^4} + 2{x^3} + c\]

Ag ionadachadh \(x = 1\) agus \(y = 3\) (bhon cho-chomharra sa cheist):

\[y = {x^4} + 2{x^3} + c\]

\[3 = {1^4} + 2{(1)^3} + c\]

\[3 = 3 + c\]

\[c = 0\]

Mar sin 's e co-aontar na lùib \(y = {x^4} + 2{x^3}\)

Eisimpleir 2 (leudachadh)

Obraich a-mach co-aontar na lùib far a bheil \(\frac{{dy}}{{dx}} = 2x + 1\) agus a' dol tron phuing \((2,9)\).

Fuasgladh

\[y = \int {(2x + 1)} \,dx = {x^2} + x + c\]

Ag ionadachadh \(x = 2\) agus \(y = 9\):

\[9 = 2^{2} + 2 + c\]

\[9 = 4 + 2 + c\]

\[c = 3\]

Mar sin 's e co-aontar na lùib \(y = {x^2} + x + 3\)

Eisimpleir 3 (leudachadh)

Tha an caisead aig tansaint ri lùb air a thoirt dhuinn mar \(f\textquotesingle(x)= 6x - \frac{5}{{{x^2}}}\). Obraich a-mach co-aontar na lùib ma tha e a' dol tron phuing \((1,6)\).

Fuasgladh

\[f(x) = \int {6x - \frac{5}{{{x^2}}}\,} dx\]

\[= \int {(6x} - 5{x^{ - 2}})dx\]

\[f(x) = \frac{{6{x^2}}}{2} - \frac{{5{x^{ - 1}}}}{{ - 1}} + c\]

\[f(x) = 3{x^2} + \frac{5}{x} + c\]

Ionadaich nuair a tha \(x = 1\) agus \(y = 6\):

\[6 = 3{(1)^2} + \frac{5}{1} + c\]

\[6 = 8 + c\]

\[c = - 2\]

Mar sin 's e co-aontar na lùib \(f(x) = 3{x^2} + \frac{5}{x} - 2\)

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