Resolved: "Evaluate the definite integral. ∫01(2t−1)2dt"

gamau6v

Answered question

2021-11-18

Evaluate the definite integral.

\int_{0}^{1} {(2 t - 1)}^{2} d t

William Yazzie

Beginner2021-11-19Added 20 answers

Step 1
Given :

\int_{0}^{1} {(2 t - 1)}^{2} d t

To evaluate the integral.
Step 2

\int_{0}^{1} {(2 t - 1)}^{2} d t

= \int_{0}^{1} (4 t^{2} - 2 (2 t) (1) + {(1)}^{2}) d t

= \int_{0}^{1} (4 t^{2} - 4 t + 1) d t

= {(4 \frac{t^{3}}{3} - 4 \frac{t^{2}}{2} + t)}_{0}^{1}

= (4 (\frac{1^{3}}{3}) - 4 (\frac{1^{2}}{2}) + 1) - (4 (\frac{0^{3}}{3}) - 4 (\frac{0^{2}}{2}) + 0)

= (\frac{4}{3} - \frac{4}{2} + 1) - 0

= (\frac{4}{3} - 2 + 1) = (\frac{4}{3} - 1) = \frac{4 - 3}{3} = \frac{1}{3}

Thus,

\int_{0}^{1} {(2 t - 1)}^{2} d t = \frac{1}{3}

Gladys Fasching

Beginner2021-11-20Added 11 answers

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:

\int_{a}^{b} f (x) d x = F (x) ∣_{a}^{b} = F (b) - F (a)

Step 2: In this case,

f (t) = {(2 t - 1)}^{2}

. Find its integral.

\frac{4 t^{3}}{3} - 2 t^{2} + t ∣_{0}^{1}

Step 3: Since

F (t) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(1)−F(0):

(\frac{4 \times 1^{3}}{3} - 2 \times 1^{2} + 1) - (\frac{4 \times 0^{3}}{3} - 2 \times 0^{2} + 0)

Step 4: Simplify.

\frac{1}{3}

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