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UPSC NDA I 2016 Mathematics Solved Paper

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Question : 80 of 120

Marks: +1, -0

If

\overset{5}{\underset{-2}{∫}}f(x)dx=4

\overset{5}{\underset{0}{∫}}(1+f(x))dx=7

\overset{0}{\underset{-2}{∫}}f(x)dx

equal to?

$-3$
2
3
5

Solution:

Concept:

Use the additive property of definite integrals:

\displaystyle\int_{a}^{c}f(x)dx = \int_{a}^{b}f(x)dx + \int_{b}^{c}f(x)dx

.

Explanation:

Given

\displaystyle\int_{-2}^{5}f(x)dx = 4

and

\displaystyle\int_{0}^{5}(1+f(x))dx = 7

.

First, simplify the second integral:

\displaystyle\int_{0}^{5}(1+f(x))dx = \int_{0}^{5}1\,dx + \int_{0}^{5}f(x)dx = [x]_{0}^{5} + \int_{0}^{5}f(x)dx = (5-0) + \int_{0}^{5}f(x)dx = 5 + \int_{0}^{5}f(x)dx = 7

.

Thus,

\displaystyle\int_{0}^{5}f(x)dx = 7-5 = 2

.

Now, use the property on

\displaystyle\int_{-2}^{5}f(x)dx = \int_{-2}^{0}f(x)dx + \int_{0}^{5}f(x)dx = 4

.

Substitute the known value:

\displaystyle\int_{-2}^{0}f(x)dx + 2 = 4

.

Therefore,

\displaystyle\int_{-2}^{0}f(x)dx = 4-2 = 2

.

Answer:

2

(Option B).

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