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Dolly Robinson

Answered question

2021-10-30

Convert the indefinite integral into definate integral using the interval [0,1]

\int (3 x^{2} - 6 x + 5) d x

Nicole Conner

Skilled2021-10-31Added 97 answers

Step 1
Convert the indefinite integral to definite integral over [0,1] as,

\int (3 x^{2} - 6 x + 5) d x \Rightarrow \int_{0} {1} (3 x^{2} - 6 x + 5) d x

Step 2
Compute integral as,

\int_{0}^{1} (3 x^{2} - 6 x + 5) d x = \int_{0}^{1} 3 x^{2} d x - \int_{0}^{1} 6 x d x + \int_{0}^{1} 5 d x

= 3 {[\frac{x^{2 + 1}}{2 + 1}]}_{0}^{1} - 6 {[\frac{x^{1 + 1}}{1 + 1}]}_{0}^{1} + {[5 x]}_{0}^{1}

= 3 {(\frac{x^{3}}{3})}_{0}^{1} - 6 {(\frac{x^{2}}{2})}_{0}^{1} + 5 {(x)}_{0}^{1}

= 3 (\frac{1}{3} - 0) - 6 (\frac{1}{2} - 0) + 5 (1 - 0)

1-3+5
=3
Step 3
Thus, the value of the integral is,

\int_{0}^{1} (3 x^{2} - 6 x + 5) d x = 3

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