∫ sqrt{6}tdt-এর সমাধান করুন

$$\int \sqrt { 6 } t d t$$

মূল্যায়ন করুন

$$\frac{\sqrt{6}t^{2}}{2}+С$$

সমাধানের ধাপগুলো

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

$$\sqrt{6}\int t\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t\mathrm{d}t with \frac{t^{2}}{2}.

$$\sqrt{6}\times \frac{t^{2}}{2}$$

সিমপ্লিফাই।

$$\frac{\sqrt{6}t^{2}}{2}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{\sqrt{6}t^{2}}{2}+С$$

w.r.t. t পার্থক্য করুন

$$\sqrt{6}t$$