Question:
Let $f \left(x\right)= \int\limits_0^{x} g (t) dt$ , where g is a non-zero even function. If $f \left(x+5\right)=g\left(x\right)$, then $ \int\limits_0^{x}f (t) dt $ equals-
Let $f \left(x\right)= \int\limits_0^{x} g (t) dt$ , where g is a non-zero even function. If $f \left(x+5\right)=g\left(x\right)$, then $ \int\limits_0^{x}f (t) dt $ equals-
Updated On: Apr 24, 2025
- $ \int\limits^5_{x+5} g (t)\, dt$
- $5 \int\limits^5_{x+5} g (t) \,dt$
- $ \int\limits^{x+5}_5 g (t) \,dt$
- $2 \int \limits ^{x+5}_5 g (t) \,dt$
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The Correct Option is A
Solution and Explanation
$f( x )=\int_{0}^{ x } g ( t ) dt$
$f(- x )=\int_{0}^{- x } g ( t ) dt$
put $t =- u$
$=-\int_{0}^{ x } g (- u ) du$
$=-\int_{0}^{x} g(u) d(u)=-f(x)$
$\Rightarrow f(- x )=-f( x )$
$\Rightarrow f( x )$ is an odd function
Also $f(5+x)=g(x)$
$f(- x )=\int_{0}^{- x } g ( t ) dt$
put $t =- u$
$=-\int_{0}^{ x } g (- u ) du$
$=-\int_{0}^{x} g(u) d(u)=-f(x)$
$\Rightarrow f(- x )=-f( x )$
$\Rightarrow f( x )$ is an odd function
Also $f(5+x)=g(x)$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
