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

The integral from t-1 to t of (1-t+x)*x is (2t-1)/2-(t(2t-1))/2+(3t^2-3t+1)/3

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integral from t-1 to t of (1-t+x)*x

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

\frac{2 t - 1}{2} - \frac{t ( 2 t - 1 )}{2} + \frac{3 t ^{2} - 3 t + 1}{3}

Solution steps

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

Expand

(1 - t + x) x : x - t x + x^{2}

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

Apply the Sum Rule:

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

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

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

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

\int_{- 1 + t}^{t} x^{2} d x = \frac{3 t ^{2} - 3 t + 1}{3}

= \frac{2 t - 1}{2} - \frac{t ( 2 t - 1 )}{2} + \frac{3 t ^{2} - 3 t + 1}{3}

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f (x) = \frac{x + 3}{x - 3}

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arctan (x^{4})

What is the integral from t-1 to t of (1-t+x)*x ?
The integral from t-1 to t of (1-t+x)*x is (2t-1)/2-(t(2t-1))/2+(3t^2-3t+1)/3