Question:
If $\frac{d}{dx} \left\{f\left(x\right)\right\} = g\left(x\right)$, then $\int\limits^{b}_{a}$ $f (x)g(x)dx$ is equal to
If $\frac{d}{dx} \left\{f\left(x\right)\right\} = g\left(x\right)$, then $\int\limits^{b}_{a}$ $f (x)g(x)dx$ is equal to
Updated On: Jun 18, 2022
- $\frac{1}{2}\left[f^{2}\left(b\right)-f^{2}\left(a\right)\right]$
- $\frac{1}{2}\left[g^{2}\left(b\right)-g^{2}\left(a\right)\right]$
- $f(b) - f(a)$
- $\frac{1}{2}\left[f\left(b^{2}\right)-f\left(a^{2}\right)\right]$
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The Correct Option is A
Solution and Explanation
$f(x)=\int g(x) d x$
$\int\limits_{a}^{b} f(x) g(x) d x=(f(x) \cdot f(x))_{a}^{b}-\int\limits_{a}^{b} g(x) \cdot f(x) d x$
$I=f^{2}(b)-f^{2}(a)$
$I=\frac{1}{2}\left(f^{2}(b)-f^{2}(a)\right)$
$\int\limits_{a}^{b} f(x) g(x) d x=(f(x) \cdot f(x))_{a}^{b}-\int\limits_{a}^{b} g(x) \cdot f(x) d x$
$I=f^{2}(b)-f^{2}(a)$
$I=\frac{1}{2}\left(f^{2}(b)-f^{2}(a)\right)$
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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.
