Evaluate the integral. \int_{2}^{3}[3x^{2}+2x+\frac{1}{x^{2}}]dx

druczekq4

druczekq4

Answered question

2021-11-20

Evaluate the integral.
23[3x2+2x+1x2]dx

Answer & Explanation

breisgaoyz

breisgaoyz

Beginner2021-11-21Added 14 answers

Step 1
The given integral is 23[3x2+2x+1x2]dx.
Step 2
Evaluate the given integral as shown below:
23[3x2+2x+1x2]dx=23[3x2]dx+23[2x]dx+23[1x2]dx
=323x2dx+223xdx+23x2dx
=3[x33]23+2[x22]23+[x11]23
=[x3]23+[x2]23[1x]23
=[278]+[94][1312]
=[19]+[5]+[16]
=1456
Rosemary McBride

Rosemary McBride

Beginner2021-11-22Added 10 answers

Step 1: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
abf(x)dx=F(x)ab=F(b)F(a)
Step 2: In this case, f(x)=(3x2+2x+1x2). Find its integral.
x3+x21x23
Step 3: Since F(x)ab=F(b)F(a), expand the above into F(3)-F(2):
(33+3213)(23+2212)
Step 4: Simplify.
1456

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