∫ (from 122 to 328) of (left(2-left(x-2right)^2right)^2-left(2-05right)^2) wrt x-এর সমাধান করুন

$$\int_{ 122 }^{ 328 } ( { \left(2- { \left(x-2 \right) }^{ 2 } \right) }^{ 2 } - { \left(2-05 \right) }^{ 2 } ) d x$$

মূল্যায়ন করুন

$$\frac{10970799276608}{15}\approx 731386618440.533333333$$

সমাধানের ধাপগুলো

\left(x-2\right)^{2} প্রসারিত করতে বাইনোমিয়াল উপপাদ্য \left(a-b\right)^{2}=a^{2}-2ab+b^{2} ব্যবহার করুন৷

$$\int _{122}^{328}\left(2-\left(x^{2}-4x+4\right)\right)^{2}-\left(2-0\times 5\right)^{2}\mathrm{d}x$$

x^{2}-4x+4 এর বিপরীত সন্ধান করতে, প্রতিটি টার্মের বিপরীত সন্ধান করুন৷

$$\int _{122}^{328}\left(2-x^{2}+4x-4\right)^{2}-\left(2-0\times 5\right)^{2}\mathrm{d}x$$

-2 পেতে 2 থেকে 4 বাদ দিন।

$$\int _{122}^{328}\left(-2-x^{2}+4x\right)^{2}-\left(2-0\times 5\right)^{2}\mathrm{d}x$$

-2-x^{2}+4x এর বর্গ

$$\int _{122}^{328}x^{4}-8x^{3}+20x^{2}-16x+4-\left(2-0\times 5\right)^{2}\mathrm{d}x$$

0 পেতে 0 এবং 5 গুণ করুন।

$$\int _{122}^{328}x^{4}-8x^{3}+20x^{2}-16x+4-\left(2-0\right)^{2}\mathrm{d}x$$

2 পেতে 2 থেকে 0 বাদ দিন।

$$\int _{122}^{328}x^{4}-8x^{3}+20x^{2}-16x+4-2^{2}\mathrm{d}x$$

2 এর ঘাতে 2 গণনা করুন এবং 4 পান।

$$\int _{122}^{328}x^{4}-8x^{3}+20x^{2}-16x+4-4\mathrm{d}x$$

0 পেতে 4 থেকে 4 বাদ দিন।

$$\int _{122}^{328}x^{4}-8x^{3}+20x^{2}-16x\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{4}-8x^{3}+20x^{2}-16x\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int -8x^{3}\mathrm{d}x+\int 20x^{2}\mathrm{d}x+\int -16x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{4}\mathrm{d}x-8\int x^{3}\mathrm{d}x+20\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}-8\int x^{3}\mathrm{d}x+20\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. -8 কে \frac{x^{4}}{4} বার গুণ করুন।

$$\frac{x^{5}}{5}-2x^{4}+20\int x^{2}\mathrm{d}x-16\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. 20 কে \frac{x^{3}}{3} বার গুণ করুন।

$$\frac{x^{5}}{5}-2x^{4}+\frac{20x^{3}}{3}-16\int x\mathrm{d}x$$

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}. -16 কে \frac{x^{2}}{2} বার গুণ করুন।

$$\frac{x^{5}}{5}-2x^{4}+\frac{20x^{3}}{3}-8x^{2}$$

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.

$$\frac{328^{5}}{5}-2\times 328^{4}+\frac{20}{3}\times 328^{3}-8\times 328^{2}-\left(\frac{122^{5}}{5}-2\times 122^{4}+\frac{20}{3}\times 122^{3}-8\times 122^{2}\right)$$

সিমপ্লিফাই।

$$\frac{10970799276608}{15}$$