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

The integral from 1 to x of 2(t-1) is x^2-2x+1

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integral from 1 to x of 2(t-1)

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

x^{2} - 2 x + 1

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{x} 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

= 2 (\int_{1}^{x} t d t - \int_{1}^{x} 1 d t)

\int_{1}^{x} t d t = \frac{x ^{2} - 1}{2}

\int_{1}^{x} 1 d t = x - 1

= 2 (\frac{x ^{2} - 1}{2} - (x - 1))

Simplify

2 (\frac{x ^{2} - 1}{2} - (x - 1)) : x^{2} - 2 x + 1

= x^{2} - 2 x + 1

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x^{\frac{5}{2}}

\frac{\partial}{\partial x} (7 ln (\frac{3 x y}{7 z}))

\int \frac{sin ^{2} ( x )}{1 - cos ( x )} d x

\frac{d}{d x} (arcsin (9 x^{5}))

\frac{8 e ^{x}}{9 e ^{x} + 5}

What is the integral from 1 to x of 2(t-1) ?
The integral from 1 to x of 2(t-1) is x^2-2x+1