Integration Questions Answers | Integration Problems with Solutions

Integration is known as the inverse process of derivatives, also called anti-derivative. There are many techniques to find integrations. In this article, we will provide a few examples of integrations with detailed solutions.

Integration Formulas

Power rule of integration: ∫xn dx = $\dfrac{x^{n+1}}{n+1}$+C Integrations of constants: ∫k dx = kx+C ∫sin x dx = -cosx+C ∫cos x dx = sin x+C ∫sec2 x dx = tan x+C ∫cosec2 x dx = -cot x+C ∫sec x tan x dx = sec x+C ∫cosec x cot x dx = -cosec x+C ∫ex dx = ex+C ∫ax dx = $\dfrac{a^x}{\ln a}$+C; a(≠1)>0. ∫$\dfrac{dx}{x}$ = ln|x| +C

Integration: Definition, Formulas, Properties, Examples

Integration Examples and Solutions

We will solve a few integrals here depending upon the type of the integrands. They will be categorized as follows:

Power Rule of Integration Examples

Question 1: Evaluate ∫6x2 dx

Answer:

By the power rule of integration,

∫6x² dx = 6 x²⁺¹/(2+1) + C

= 6 x³/3 + C

= 2x³ + C, where C is an integration constant.

Question 2: Evaluate ∫(x2+x+1) dx

Answer:

∫(x²+x+1) dx = ∫x² dx + ∫x dx + ∫dx

= x³/3 + x²/2 + x + C

= 2x³ + C, where C is an integration constant.

Question 2: Evaluate ∫1/x2 dx

Answer:

Note that 1/x² can be written as x^-2 by the rule of indices. We have:

∫1/x² dx = ∫x^-2 dx

= x^-2+1/(-2+1) + C

= x^-1/-1 + C

= -1/x + C, where C is an integration constant.

Trigonometric Integration Examples

Question 4: Evaluate ∫sin 2x dx

Answer:

We will use the substitution method to integrate. Let us put

z=2x ⇒ dz=2 dx ⇒dx=dz/2.

Thus, ∫sin 2x dx = ∫sin z dz/2

= (1/2)∫sin z dz = (1/2) (-cos z) + C

= $-\dfrac{\cos 2x}{2}$ + C as z=2x.

Also Read:

Integration of mod x

Integration of Root x

Question 5: Evaluate ∫cos2x dx

Answer:

We will use the following trigonometric formula: cos 2x = 2cos²x -1. Thus, we have

cos²x = $\dfrac{1+\cos 2x}{2}$

∴ ∫cos²x dx = $\dfrac{1}{2} \int (1+\cos 2x) dx$

= $\dfrac{1}{2} (x+\dfrac{\sin 2x}{2})$ as we know that ∫cos mx dx = $\dfrac{\sin mx}{m}$ +C

= $\dfrac{x}{2} +\dfrac{\sin 2x}{4}$

Exponential Integration Examples

Question 6: Evaluate ∫e2x dx

Answer:

Put 2x=z. so we have dx=dz/2.

∴ ∫e^2x dx = $\int e^z \dfrac{dz}{2}$

= $\dfrac{1}{2} \int e^z dz$

= $\dfrac{1}{2} e^z$ +C

= $\dfrac{1}{2} e^{2x}$ +C as z=2x.

Question 7: Evaluate ∫(ex+e-x) dx

Answer:

∫(e^x+e^-x) dx

= ∫e^x dx + ∫e^-x dx

= e^x – e^-x +C as we know that ∫e^mx dx = $\dfrac{e^{mx}}{m}$ +C.

Question 8: Evaluate ∫2x dx

Answer:

We know that ∫a^x dx = $\dfrac{a^x}{\ln a}$ +C; a(≠1)>0. Here ln denotes the natural logarithm, that is, the logarithm with base e.

So applying the above formula for a=2, we obtain that

∫2^x dx = $\dfrac{2^x}{\ln 2}$+C.

Logarithmic Integration Examples

Question 9: Evaluate ∫logx dx

Answer:

To find the integration of logx, we will use the integration by parts formula:

∫uv dx = u∫v dx $-\int [\frac{du}{dx}( \int v \, dx )]\, dx$, where u, v are functions of x.

Put u=logx, v=1

∴ ∫logx dx

= ∫(logx × 1) dx

= log x ∫dx – ∫x [d/dx(logx)] dx +C

= xlog x – ∫x (1/x) dx +C

= xlog x – ∫dx +C

= xlog x – x +C

So the integration of logx is equal to x(logx-1)+C.

Question 10: Evaluate ∫$\dfrac{\log x}{x}$ dx

Answer:

Put z=logx

So dz = dx/x.

∴ ∫$\dfrac{\log x}{x}$ dx

= ∫z dz

= z²/2 +C

= (log x)²/2 +C, where C is an integral constant.