Question

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HEY PUBLIC

SOLVE INTEGRATION..

100 points... BRAINLIST ​

Answer 1

Answer: 6

Step-by-step explanation:

Let,

$I=\int\limits^{4}_{-2}{x}\,dx\\\;\\\;=[\frac{x^2}{2}]\limits^{4}_{-2}\\\;\\=\frac{1}{2}[x^2]\limits^{4}_{-2}\\\;\\\text{Putting Limits}\\\;\\=\frac{1}{2}[4^2-(-2)^2]\\\;\\=\frac{1}{2}[16-4]\\\;\\=\frac{1}{2}\times12\\\;\\=6$

Note:- $\int x^n\;dx=\frac{x^{n+1}}{n+1}$

Answer 2

$\mathfrak{\huge{\green{\underline{\underline{AnswEr :}}}}}$

⠀ 4

⇒∫ x Dx

⠀-2

⇒(x²/2)⁴ -2

⇒1/2 (x²)⁴ -2

⇒ 1/2 [ 4² - (-2)² ]

⇒ 1/2 [ 4×4 - (-2×-2) ]

⇒ 1/2 [ 16 - 4 ]

⇒ 1/2 × 12

⇒6

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