Definite Integration
Definite Integration
nta_pyq_2025_apr
Grade 12

Question:

Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x) = \displaystyle\int_0^x tf(t)\,dt$. If $g(x^3) = x^6+x^7$, then value of $\displaystyle\sum_{r=1}^{15}f(r^3)$ is:
$270$
$340$
$320$
$310$

Step-by-Step Solution

Key Concept: Differentiate $g(x^3) = x^6+x^7$ with respect to $x$ using the chain rule to get $g'(x^3)\cdot 3x^2 = 6x^5+7x^6$, then use $g'(u) = uf(u)$ to extract $f(u)$.
Since $g(x) = \int_0^x tf(t)dt$, by FTC $g'(x) = xf(x)$. Differentiating $g(x^3) = x^6+x^7$: $$g'(x^3)\cdot 3x^2 = 6x^5+7x^6 \Rightarrow x^3 f(x^3)\cdot 3x^2 = 6x^5+7x^6.$$ $$3x^5 f(x^3) = 6x^5+7x^6 \Rightarrow f(x^3) = 2+\frac{7x}{3}.$$ So $f(r^3) = 2+\dfrac{7r}{3}$. $$\sum_{r=1}^{15}f(r^3) = 2(15)+\frac{7}{3}\cdot\frac{15\cdot 16}{2} = 30+280 = 310.$$
Correct Answer: 4

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