Expert answers to "Evaluate the definity integral. ∫04(x−3)(x−5)dx"

jeliceg2

Answered question

2021-11-22

Evaluate the definity integral.

\int_{0}^{4} (x - 3) (x - 5) d x

Mollicchiuk

Beginner2021-11-23Added 13 answers

Step 1
Evaluate the integral

\int_{0}^{4} (x - 3) (x - 5) d x

.
Step 2

\int_{0}^{4} (x - 3) (x - 5) d x = \int_{0}^{4} x^{2} - 8 x + 15 d x

= {(\frac{x^{3}}{3} - 4 x^{2} + 15 x)}_{0}^{4}

= (\frac{4^{3}}{3} - 4^{3} + 15 \times 4)

= \frac{52}{3}

John Timms

Beginner2021-11-24Added 8 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(x)=(x-3)(x-5). Find its integral.

\frac{x^{3}}{3} - 4 x^{2} + 15 x ∣_{0}^{4}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(4)−F(0):

(\frac{4^{3}}{3} - 4 \times 4^{2} + 15 \times 4) - (\frac{0^{3}}{3} - 4 \times 0^{2} + 15 \times 0)

Step 4: Simplify.

= \frac{52}{3}

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