Suppose $f$ is such that $f( - x) = - f(x)$ for every real $x$ and $\int_{0}^{1}{f(x)dx = 5},$ then $\int_{- 1}^{0}{f(t)dt}\$is equal to

Question

Dr. Abubacker · Accepted Answer

(d)
Given, $f( - x) = - f(x),\forall\$values of real $x$
We know that,
$\int_{- a}^{a}{f(x)dx = 0 = \int_{- a}^{0}{f(x)dx +}\int_{0}^{a}{f(x)dx}}\ \ \ \lbrack\because f( - x) = - f(x)\rbrack$
$\Rightarrow \ \int_{- 1}^{0}{f(x)}dx + \int_{0}^{1}{f(x)dx = 0}$
$\Rightarrow \ \int_{- 1}^{0}{f(x)dx = - 5}\ \ \lbrack\because\ \int_{0}^{1}{f(x)dx = 5\rbrack}$
$\Rightarrow \ \int_{- 1}^{0}{f(t)dt = - 5}$

Q. Suppose $f$ is such that $f( - x) = - f(x)$ for every real $x$ and $\int_{0}^{1}{f(x)dx = 5},$ then $\int_{- 1}^{0}{f(t)dt}\$is equal to

Explanation

Reviewed by Dr. Abubacker

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