Question

$\displaystyle\sf\int\limits_{2}^{8}(4x + 6)dx$​

Answer 1

$\\ \rm\rightarrowtail {\displaystyle\int\limits_{2}^{8}}(4x+6)dx$

$\\ \rm\rightarrowtail [2x^2]{}_2^8+[6x]{}_2^8$

$\\ \rm\rightarrowtail [2(8)^2-2(2)^2]+[6(8)-6(2)]$

$\\ \rm\rightarrowtail [2(64)-2(4)]+[48-12]$

$\\ \rm\rightarrowtail 128-8+36$

$\\ \rm\rightarrowtail 120+36$

$\\ \rm\rightarrowtail 156$

Answer 2

The result obtained by carrying out the integration of ₂∫⁸(4x + 6)dx is 156

Integral calculus

if y = xⁿ

Then the integration of y will be

∫ydx = [x^(n + 1)] / n + 1

Data obtained from the question

• ₂∫⁸(4x + 6)dx
• Integration =?

How to integrate ₂∫⁸(4x + 6)dx

We shall integrate individually

∫4x = 4x² / 2 = 2x²

∫6 = 6x

Thus, the integration of (4x + 6) is given as

∫4x + 6 = 2x² + 6x

Now, we shall consider the limits (i.e 8 and 2)

∫(4x + 6)dx = 2x² + 6x

₂∫⁸(4x + 6)dx = [2(8)² + 6(8)] – [2(2)² + 6(2)]

₂∫⁸(4x + 6)dx = [2(64) + 48)] – [2(4) + 12]

₂∫⁸(4x + 6)dx = [128 + 48] – [8 + 12]

₂∫⁸(4x + 6)dx = 176 – 20

₂∫⁸(4x + 6)dx = 156

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