∫ (from 0 to 04) of -3x-(sqrt{x}) wrt x-এর সমাধান করুন

$$\int _ { 0 } ^ { 04 } - 3 x - ( \sqrt { x } ) d x$$

মূল্যায়ন করুন

$$0$$

সমাধানের ধাপগুলো

Evaluate the indefinite integral first.

$$\int -3x-\sqrt{x}\mathrm{d}x$$

Integrate the sum term by term.

$$\int -3x\mathrm{d}x+\int -\sqrt{x}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-3\int x\mathrm{d}x-\int \sqrt{x}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. -3 কে \frac{x^{2}}{2} বার গুণ করুন।

$$-\frac{3x^{2}}{2}-\int \sqrt{x}\mathrm{d}x$$

x^{\frac{1}{2}} হিসেবে \sqrt{x} পুনরায় লিখুন৷ Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{3}{2}}}{\frac{3}{2}}. সিমপ্লিফাই। -1 কে \frac{2x^{\frac{3}{2}}}{3} বার গুণ করুন।

$$-\frac{3x^{2}}{2}-\frac{2x^{\frac{3}{2}}}{3}$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$-\frac{3}{2}\times \left(0\times 4\right)^{2}-\frac{2}{3}\times \left(0\times 4\right)^{\frac{3}{2}}-\left(-\frac{3}{2}\times 0^{2}-\frac{2}{3}\times 0^{\frac{3}{2}}\right)$$

সিমপ্লিফাই।

$$\text{Indeterminate}$$