What is the integral from-3 to x of (t+3) ?

The integral from-3 to x of (t+3) is (x^2-9)/2+3x+9

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integral from-3 to x of (t+3)

\int_{- 3}^{x} (t + 3) d t

\frac{x ^{2} - 9}{2} + 3 x + 9

Solution steps

\int_{- 3}^{x} t + 3 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_{- 3}^{x} t d t + \int_{- 3}^{x} 3 d t

\int_{- 3}^{x} t d t = \frac{x ^{2} - 9}{2}

\int_{- 3}^{x} 3 d t = 3 x + 9

= \frac{x ^{2} - 9}{2} + 3 x + 9

\int \frac{cos ( x )}{sin ^{5} ( x )} d x

f (x) = \frac{x ^{3} - 3 x ^{2} + 4}{x ^{2}}

\frac{d}{d x} (e^{x} \cdot cos (x))

\frac{\partial}{\partial x} (6 x e^{x + y})

x \to 3 - lim (2 3 - x)

What is the integral from-3 to x of (t+3) ?
The integral from-3 to x of (t+3) is (x^2-9)/2+3x+9