2,1(1x3+2) find the integral

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Froldigh · Accepted Answer

Step 1  To Determine: find the integral of (1x3+2) in interval (2,1)  Given: we have a function f(x)=y=(1x3+2). Also we have limits 2 to 1  Explanation: we have an integral   ∫21(1x3+2)dx  so we will integrate this as follows  Step 2  ∫21(x−3+2)dx=[x−3+1−3+1+2x]12  =[x−2−2+2x]21  =[−12x2+2x]21  taking limits we have   ∫21(x−3+2)dx=[−12x2+2x]21  [(−12×12+2×1)−(−12×22+2×2)]  =[(−12+2)−(−18+4)]  =[−12+2+18−4]  =−38−2  =−198

inenge3y · Accepted Answer

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) = (\frac{1}{x^{3}} + 2)

. Find its integral.

- \frac{1}{2 x^{2}} + 2 x ∣_{2}^{1}

Step 3: Since

F (x) ∣_{a}^{b} = F (b) - F (a)

, expand the above into F(1)-F(2):

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

Step 4: Simplify.

- \frac{19}{8}

2,1 (1/x^{3}+2) find the integral

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