What is the integral from x to x+2 of (4t+1) ?

The integral from x to x+2 of (4t+1) is 8x+10

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integral from x to x+2 of (4t+1)

\int_{x}^{x + 2} (4 t + 1) d t

8 x + 10

Solution steps

\int_{x}^{x + 2} 4 t + 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{x}^{x + 2} 4 t d t + \int_{x}^{x + 2} 1 d t

\int_{x}^{x + 2} 4 t d t = 8 (x + 1)

\int_{x}^{x + 2} 1 d t = 2

= 8 (x + 1) + 2

Simplify

8 (x + 1) + 2 : 8 x + 10

= 8 x + 10

\frac{\partial}{\partial x} (2 cos (x y))

y = 2 x^{3} - 5 x, (1, - 3)

f (x) = 2 x^{3}, (- 1, - 2)

\frac{\partial}{\partial x} (\frac{1}{2} sin (2 x - y))

\frac{\partial}{\partial x} (\frac{x ^{3} b ^{4}}{c ^{2} d ^{4}})

What is the integral from x to x+2 of (4t+1) ?
The integral from x to x+2 of (4t+1) is 8x+10