If p(x) is a cubic polynomial with p(1) = 3, p(0) = 2 and p(–1) = 4, then \int ^{1} _{–1} p(x)dx is -

KVPY SX SB Exam 2015 Question Paper

Question : 13 of 120

Marks: +1, -0

If p(x) is a cubic polynomial with p(1) = 3, p(0) = 2 and p(–1) = 4, then

∫ ↖{1} ↙{–1} p(x)dx

is -

Solution:

Let

P(x) = ax^3 + bx^2 + cx + d

a + b + c + d = 3
d = 2
– a + b – c + d = 4
2b + 2d = 7
2b + 4 = 7
2b = 3

b=3/2

∫ ↖{1} ↙{–1} (ax^3+ bx^2+ cx+ d)dx

2∫ ↖{1} ↙{–1} (bx^2+ d)dx

=2[bx^3/3+dx]_0^1

=2(b/3+d)

=2(1/2+2)=5

Go to Question:

Prev Question

Next Question