# Definite Integral Problems and Solutions [SLST 2016]

In this post, we will discuss a few objective questions on definite integrals with answers. This will help us to take preparation for various exams, like board exams, SLST, and many more.

Q1: Find $\int_0^\pi$|cosx| dx. [WB SLST 16]

We know that cosx is positive for x∈ [0, π/2] and cosx has negative values in [π/2, π]. Thus, we will proceed as follows:

$\int_0^\pi$|cosx| dx

= $\int_0^{\pi/2}$|cosx| dx + $\int_{\pi/2}^\pi$|cosx| dx

= $\int_0^{\pi/2}$cosx dx – $\int_{\pi/2}^\pi$ cosx dx

= $[\sin x]_0^{\pi/2} – [\sin x]_{\pi/2}^\pi$

= sin $\frac{\pi}{2}$ – sin0 – sinπ + sin $\frac{\pi}{2}$

= 1 -0 -0 +1

= 2.

So the definite integral of mod cosx from 0 to π is equal to 2.

Q2: Find the value of $\int_2^4$[|x-1| + |x-3|] dx. [WB SLST 16]

$\int_2^4$[|x-1| + |x-3|] dx

= $\int_2^4$ (x-1) dx + $\int_2^4$ |x-3| dx, as x-1 is always positive in [2, 4].

= $\int_2^4$ (x-1) dx + $\int_2^3$ |x-3| dx + $\int_3^4$ |x-3| dx

= $\int_2^4$ (x-1) dx – $\int_2^3$ (x-3) dx + $\int_3^4$ (x-3) dx

= [x2/2 -x]$_2^4$ – [x2/2 -3x]$_2^3$ + [x2/2 -3x]$_3^4$

= (42/2 -4) – (22/2 -2) – (32/2 -9) + (22/2 -6) + (42/2 -12) – (32/2 -9)

= (8 -4) – (2 -2) – (9/2 -9) + (2 -6) + (8 -12) – (9/2 -9)

= 4 -0 -2(9/2 -9) -4 -4

= -4 -9+18

= 5

Q3: Find the value of $\int_{-\pi}^\pi \dfrac{\cos^2 x}{1+a^x}$, where a>0. [WB SLST 16]

Let I = $\int_{-\pi}^\pi \dfrac{\cos^2 x}{1+a^x}$ …(A)

If we change the variable x to -x, then the integral will remain the same. So

I = $\int_{-\pi}^\pi \dfrac{\cos^2 (-x)}{1+a^{-x}}$

⇒ I = $\int_{-\pi}^\pi a^x \dfrac{\cos^2 x}{1+a^{x}}$ as cos(-x)=cosx …(B)

Adding (A) and (B), we get that

2I = $\int_{-\pi}^\pi$ cos2x dx

= $\int_{-\pi}^\pi \dfrac{1+\cos 2x}{2}$ dx

= $\int_{0}^\pi$ (1+cos 2x) dx

= $[x+\frac{\sin 2x}{2}]_0^\pi$

= (π + sin2π) – (0+sin0)

= π

⇒ I = π/2.