# Gate 2015 – Set 1 – EE Question 1 – Solution

Gate 2015 – Set 1 – EE

Question 1 – Solution

$$\int_{-\infty }^{\infty }f(x) dx =1 \Rightarrow \int_{0}^{1}(a+bx)dx =1 \Rightarrow \left [ax +\frac{bx^{2}}{2} \right ]_{0}^{1}=1$$