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2015 Paper 1 Q5
D: 1516.0
B: 1500.0
integration
curve sketching
definite integral
substitution
rational function
square root
parameter as variable
- The function \(\f\) is defined, for \(x>0\), by \[ \f(x) =\int_{1}^3 (t-1)^{x-1} \, \d t \,. \] By evaluating the integral, sketch the curve \(y=\f(x)\).
- The function \(\g\) is defined, for \(-\infty < x < \infty\), by \[ \g(x)= \int_{-1}^1 \frac 1 {\sqrt{1-2xt +x^2} \ }\, \d t \,.\] By evaluating the integral, sketch the curve \(y=\g(x)\).
Solution:
- \(\,\)
\begin{align*}
&& f(x) &= \int_1^3 (t-1)^{x-1} \d t \\
&&&= \left [ \frac1x(t-1)^{x} \right]_1^3 \\
&&&= \frac{2^x}{x}
\end{align*}
- \(\,\)
\begin{align*}
&& g(x) &= \int_{-1}^1 \frac{1}{\sqrt{1-2xt+x^2}} \d t \\
&&&= \left [ -\frac{1}{x}(1 +x^2 - 2xt)^{\frac12} \right]_{-1}^1 \\
&&&= \frac1x \left ( \sqrt{1+x^2+2x}-\sqrt{1+x^2-2x}\right) \\
&&&= \frac1x \left ( |1+x|-|1-x| \right)
\end{align*}
