{"id":32285,"date":"2025-06-22T12:21:59","date_gmt":"2025-06-22T12:21:59","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32285"},"modified":"2025-06-22T12:22:01","modified_gmt":"2025-06-22T12:22:01","slug":"integrate","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/integrate\/","title":{"rendered":"Integrate."},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are asked to evaluate the definite integral:\u222b15ln\u2061(3x)\u2009dx\\int_1^5 \\ln(3x) \\, dx\u222b15\u200bln(3x)dx<\/p>\n\n\n\n

\n\n\n\n

Step 1: Use the logarithmic identity<\/h3>\n\n\n\n

Recall the identity:ln\u2061(3x)=ln\u20613+ln\u2061x\\ln(3x) = \\ln 3 + \\ln xln(3x)=ln3+lnx<\/p>\n\n\n\n

So we can rewrite the integral as:\u222b15ln\u2061(3x)\u2009dx=\u222b15(ln\u20613+ln\u2061x)\u2009dx\\int_1^5 \\ln(3x) \\, dx = \\int_1^5 (\\ln 3 + \\ln x) \\, dx\u222b15\u200bln(3x)dx=\u222b15\u200b(ln3+lnx)dx<\/p>\n\n\n\n

Now break it into two separate integrals:=\u222b15ln\u20613\u2009dx+\u222b15ln\u2061x\u2009dx= \\int_1^5 \\ln 3 \\, dx + \\int_1^5 \\ln x \\, dx=\u222b15\u200bln3dx+\u222b15\u200blnxdx<\/p>\n\n\n\n

Since ln\u20613\\ln 3ln3 is a constant:=ln\u20613\u222b15dx+\u222b15ln\u2061x\u2009dx= \\ln 3 \\int_1^5 dx + \\int_1^5 \\ln x \\, dx=ln3\u222b15\u200bdx+\u222b15\u200blnxdx<\/p>\n\n\n\n

The first integral is straightforward:\u222b15dx=5\u22121=4\\int_1^5 dx = 5 – 1 = 4\u222b15\u200bdx=5\u22121=4<\/p>\n\n\n\n

So:ln\u20613\u22c54=4ln\u20613\\ln 3 \\cdot 4 = 4 \\ln 3ln3\u22c54=4ln3<\/p>\n\n\n\n

\n\n\n\n

Step 2: Compute \u222b15ln\u2061x\u2009dx\\int_1^5 \\ln x \\, dx\u222b15\u200blnxdx<\/h3>\n\n\n\n

Use integration by parts:<\/p>\n\n\n\n

Let
u=ln\u2061x\u21d2du=1xdxu = \\ln x \\Rightarrow du = \\frac{1}{x} dxu=lnx\u21d2du=x1\u200bdx
dv=dx\u21d2v=xdv = dx \\Rightarrow v = xdv=dx\u21d2v=x<\/p>\n\n\n\n

So:\u222bln\u2061x\u2009dx=xln\u2061x\u2212\u222bx\u22c51xdx=xln\u2061x\u2212\u222b1dx=xln\u2061x\u2212x+C\\int \\ln x \\, dx = x \\ln x – \\int x \\cdot \\frac{1}{x} dx = x \\ln x – \\int 1 dx = x \\ln x – x + C\u222blnxdx=xlnx\u2212\u222bx\u22c5x1\u200bdx=xlnx\u2212\u222b1dx=xlnx\u2212x+C<\/p>\n\n\n\n

Now evaluate the definite integral:\u222b15ln\u2061x\u2009dx=[xln\u2061x\u2212x]15\\int_1^5 \\ln x \\, dx = \\left[ x \\ln x – x \\right]_1^5\u222b15\u200blnxdx=[xlnx\u2212x]15\u200b<\/p>\n\n\n\n

Evaluate at the limits:<\/p>\n\n\n\n

At x=5x = 5x=5:
5ln\u20615\u221255 \\ln 5 – 55ln5\u22125<\/p>\n\n\n\n

At x=1x = 1x=1:
1ln\u20611\u22121=0\u22121=\u221211 \\ln 1 – 1 = 0 – 1 = -11ln1\u22121=0\u22121=\u22121<\/p>\n\n\n\n

So:\u222b15ln\u2061x\u2009dx=(5ln\u20615\u22125)\u2212(\u22121)=5ln\u20615\u22124\\int_1^5 \\ln x \\, dx = (5 \\ln 5 – 5) – (-1) = 5 \\ln 5 – 4\u222b15\u200blnxdx=(5ln5\u22125)\u2212(\u22121)=5ln5\u22124<\/p>\n\n\n\n

\n\n\n\n

Step 3: Combine everything<\/h3>\n\n\n\n

\u222b15ln\u2061(3x)\u2009dx=4ln\u20613+(5ln\u20615\u22124)\\int_1^5 \\ln(3x) \\, dx = 4 \\ln 3 + (5 \\ln 5 – 4)\u222b15\u200bln(3x)dx=4ln3+(5ln5\u22124)=4ln\u20613+5ln\u20615\u22124= 4 \\ln 3 + 5 \\ln 5 – 4=4ln3+5ln5\u22124<\/p>\n\n\n\n

\n\n\n\n

Step 4: Approximate<\/h3>\n\n\n\n

Use calculator values:<\/p>\n\n\n\n

    \n
  • ln\u20613\u22481.0986\\ln 3 \\approx 1.0986ln3\u22481.0986<\/li>\n\n\n\n
  • ln\u20615\u22481.6094\\ln 5 \\approx 1.6094ln5\u22481.6094<\/li>\n<\/ul>\n\n\n\n

    So:4ln\u20613\u22484\u22c51.0986=4.39444 \\ln 3 \\approx 4 \\cdot 1.0986 = 4.39444ln3\u22484\u22c51.0986=4.39445ln\u20615\u22485\u22c51.6094=8.04705 \\ln 5 \\approx 5 \\cdot 1.6094 = 8.04705ln5\u22485\u22c51.6094=8.04704.3944+8.0470\u22124=8.44144.3944 + 8.0470 – 4 = 8.44144.3944+8.0470\u22124=8.4414<\/p>\n\n\n\n

    \n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    \u222b15ln\u2061(3x)\u2009dx\u22488.4414\\int_1^5 \\ln(3x) \\, dx \\approx \\boxed{8.4414}\u222b15\u200bln(3x)dx\u22488.4414\u200b<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The Correct Answer and Explanation is: We are asked to evaluate the definite integral:\u222b15ln\u2061(3x)\u2009dx\\int_1^5 \\ln(3x) \\, dx\u222b15\u200bln(3x)dx Step 1: Use the logarithmic identity Recall the identity:ln\u2061(3x)=ln\u20613+ln\u2061x\\ln(3x) = \\ln 3 + \\ln xln(3x)=ln3+lnx So we can rewrite the integral as:\u222b15ln\u2061(3x)\u2009dx=\u222b15(ln\u20613+ln\u2061x)\u2009dx\\int_1^5 \\ln(3x) \\, dx = \\int_1^5 (\\ln 3 + \\ln x) \\, dx\u222b15\u200bln(3x)dx=\u222b15\u200b(ln3+lnx)dx Now break it into […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32285","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32285","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=32285"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32285\/revisions"}],"predecessor-version":[{"id":32288,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32285\/revisions\/32288"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32285"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32285"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}