   Chapter 7.4, Problem 64E

Chapter
Section
Textbook Problem

# Find the area of the region under the given curve from 1 to 2. y = 1 x 3 + x

To determine

To find: Find the area of the region under the given curve y=1x3+x

Explanation

Calculation: The area under the curve y=1x3+x from x=1 to x=2 is A=abydx

121x3+xdx=121x(x2+1)dx

Since x2+1 has no real roots we decompose the function into partial fractions

Ax+Bx+Cx2+1=1x3+x. Then, A(x2+1)+(Bx+C)xx3+x=1x3+x

Thus we want that A(x2+1)+(Bx+C)x=1

Taking x=0 we have that A(02+1)=1. Hence A=1

Taking x=1 we have that A(12+1)+(B.1+C).1=1. Hence 2A+B+C=1. Since A=1 we have that B+C=1

Taking x=1 we have that A((1)2+1)+(B.(1)+C)(1)=1. Hence 2A+BC=1. Since A=1 we have that BC=1

Thus we have that B+C=BC which gives us that C=C

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