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.

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

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

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 $$y = {x^4} + 2{x^3}$$

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

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

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 $$y = {x^2} + x + 3$$

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 $$(1,6)$$.

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

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

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

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

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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