Mathematics Integral Calculus Application of Integrals

The area bounded by the curves y = –(x + 1)2 ; y = |x + 1| and |y| = 1 – x is

1
\(\frac{11}{6}\)
2
\(\frac{13}{6}\)
3
\(\frac{9}{6}\)
4
\(\frac{3}{4}\)

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