If int_{0}^{1} f(x) dx = 5,then the value of | vedclass

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(D) Given,$\int_{0}^{1} f(x) dx = 5$. Let $I = 100 \int_{0}^{1} x^{9} f(x^{10}) dx$. Substitute $x^{10} = t$,then $10x^{9} dx = dt$,which implies $x^{

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$55$

Vedclass · Answer

(D) Given,$\int_{0}^{1} f(x) dx = 5$. Let $I = 100 \int_{0}^{1} x^{9} f(x^{10}) dx$. Substitute $x^{10} = t$,then $10x^{9} dx = dt$,which implies $x^{9} dx = \frac{dt}{10}$. When $x = 0, t = 0$ and when $x = 1, t = 1$. Therefore,$I = 100 \int_{0}^{1} f(t) \frac{dt}{10} = 10 \int_{0}^{1} f(t) dt$. Since $\int_{0}^{1} f(t) dt = \int_{0}^{1} f(x) dx = 5$,we have $I = 10 	imes 5 = 50$. Thus,the required value is $\int_{0}^{1} f(x) dx + I = 5 + 50 = 55$.

If $\int_{0}^{1} f(x) dx = 5$,then the value of $\int_{0}^{1} f(x) dx + 100 \int_{0}^{1} x^{9} f(x^{10}) dx$ is equal to

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