If $f(-x)=-f(x) \forall x \in K$ and $\int_0^1 f(x) d x=5$ then $\int_{-1}^0 f(t) d t=$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Medium

1 of 1

Application +4 | -1

f(-x)=-f(x) \forall x \in K

and

\int_0^1 f(x) d x=5

then

\int_{-1}^0 f(t) d t=

View answer and explanation

Correct option: D

-5

Option D (-5) is correct because the integral of an odd function over a symmetric interval is zero, and using the given integral from 0 to 1, we can find the integral from -1 to 0.

More Questions from DEFINITE INTEGRATION

Practice All Chapter MCQs →

\int \nolimits_0^\infty \,\,\log \,\,\left( {x + \frac{1}{x}} \right)\,\,\frac{{dx}}{{1 + {x^2}}}

is equal to If

f\left( x \right)

is differentiable and

\int \nolimits_0^{{t^2}} \,\,x\;\,f\left( x \right)\,\,dx = \frac{2}{5}\,\,{t^5}

, then

f\left( {\frac{4}{{25}}} \right)

equals The value of

\int\limits_{{0}}^{11}[x]^{3} . \mathrm{dx}

, where

[*]

denotes the greatest integer function, is The integral

\int\limits_{2}^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x

is equal to: The definite integral

\int\limits_{{-2}}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x

equals If

I\,\,\left( {m,\,\,n} \right) = \int \nolimits_0^1 \,\,{t^m}\,\,{\left( {1 + t} \right)^n}\,\,dt

,then the expression for

I\left( {m,\,\,n} \right)

in terms of

I\,\,\left( {m + 1,\,\,n + 1} \right)