Solve the following Question.(1 Marks)

Question 11 Mark

$\int_{-2}^3 \frac{1}{x+5} d x$

Answer

$
\begin{aligned}
& \text { Let } I =\int_{-2}^3 \frac{1}{x+5} \cdot d x \\
& =[\log |x+5|]_{-2}^3 \\
& =[\log |3+5|-\log |-2+5|] \\
& =\log 8-\log 3 \\
& \therefore I =\log \left(\frac{8}{3}\right) .
\end{aligned}
$

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Question 21 Mark

$\int_4^9 \frac{1}{\sqrt{x}} d x$

Answer

$
\begin{aligned}
& \text { Let } I =\int_4^9 \frac{1}{\sqrt{x}} \cdot d x \\
& =\int_4^9 x^{\frac{1}{2}} \cdot d x=\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^9 \\
& =2[\sqrt{x}]_4^9 \\
& =2(\sqrt{9}-\sqrt{4}) \\
& =2(3-2) \\
& \therefore I =2 .
\end{aligned}
$

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Question 31 Mark

$\int_0^1 \frac{1}{2 x-3} d x$

Answer

$
\begin{aligned}
& \int_0^1 \frac{1}{2 x-3} d x=\left[\frac{\log |2 x-3|}{2}\right]_0^1 \\
& =\frac{1}{2}[\log |-1|-\log |-3|] \\
& =\frac{1}{2}(\log 1-\log 3) \\
& =-\frac{1}{2} \log 3 \text {. } \\
& \ldots[\because \log 1=0] \\
\end{aligned}
$

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Question 41 Mark

$
\int_{-4}^{-1} \frac{1}{x} d x
$

Answer

$
\begin{aligned}
& \int_{-4}^{-1} \frac{1}{x} d x=[\log |x|]_{-4}^{-1} \\
= & \log |-1|-\log |-4| \\
= & \log 1-\log 4 \\
= & -\log 4 . \quad \ldots[\because \log 1=0]
\end{aligned}
$

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Question 51 Mark

$\int_1^2 x^2 d x$

Answer

$
\begin{aligned}
\int_1^2 x^2 d x & =\left[\frac{x^3}{3}\right]_1^2 \\
& =\frac{1}{3}(8-1)=\frac{7}{3} .
\end{aligned}
$

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Question 61 Mark

Evaluate the following definite integrals : $\int_{-2}^3 \frac{1}{x+5} d x$

Answer

$
\begin{aligned}
& \int_{-2}^3 \frac{1}{x+5} d x \\
& =[\log |x+5|]_{-2}^3
\end{aligned}
$
$
\begin{aligned}
& =\log 8-\log 3 \\
& =\log \left(\frac{8}{3}\right)
\end{aligned}
$

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Question 71 Mark

Evaluate the following definite integrals : $\int_4^9 \frac{1}{\sqrt{x}} d x$

Answer

$
\begin{aligned}
& \int_4^9 \frac{1}{\sqrt{x}} d x=\int_4^9 x^{-\frac{1}{2}} d x \\
= & {\left[\frac{x^{\frac{1}{2}}}{1 / 2}\right]_4^9=2[\sqrt{x}]_4^9 } \\
= & 2(\sqrt{9}-\sqrt{4}) \\
= & 2(3-2)=2
\end{aligned}
$

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Maths (commerce) Solve the following Question.(1 Marks)

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