AP EAMCET · Maths · Definite Integration
The slope of the tangent to the curve \(y=\int_0^x \frac{1}{1+t^3} d t\) at the point, where \(x=1\) is
- A \(\frac{1}{4}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{2}\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(y=\int_0^x \frac{1}{1+t^3} d t\) Differentiating w.r.t. \(x\), we get \(\frac{d y}{d x}=\frac{1}{1+x^3}\) \(\begin{aligned} & \text { At } x=1 \\ & \therefore \quad\left(\frac{d y}{d x}\right)_{x=1}=\frac{1}{1+1}=\frac{1}{2} \end{aligned}\)
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