integral from 0 to 1 of 9x — step-by-step solution

Let's integrate:

\int_{0}^{1} 9 x d x

ⓘ A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1:

\int 9 x d x

9 \cdot \frac{x^{2}}{2}

9 \cdot \frac{x^{2}}{2}

ⓘ Multiplying a fraction by a number multiplies its numerator:

9 \cdot \frac{x^{2}}{2}

\frac{9 x^{2}}{2}

▷

\frac{9 x^{2}}{2}

Let's evaluate for the interval

[0, 1]

\frac{9 \cdot 1^{2}}{2} - \frac{9 \cdot 0^{2}}{2}

ⓘ A product with a zero factor is zero:

9 \cdot 0

0

\frac{9}{2} + 0

ⓘ Adding 0 changes nothing:

\frac{9}{2} + 0

\frac{9}{2}

▶Integral:

\frac{9}{2}

The numeric value of:

\frac{9}{2}

is:

▷

4.5